Coordinate Geometry

4.10 Coordinate Geometry Booster

24. If 3x + 4y + l = 0 is a tangent to the parabola y2 = 12x, 47. Prove that the orthocentre of any triangle formed by
find the value of l.
three tangents to a parabola lies on the directrix.
25. Find the length of the chord intercepted by the parabola
y2 = 4ax and the line y = mx + c. 48. Prove that the equation of the director circle to the
parabola y2 = 4ax is x + a = 0.
TANGENT AND TANGENCY
49. Find the equation of the director circle to the following
26. Find the point of intersection of tangents at P(t1) and
Q(t2) on the parabola y2 = 4ax. parabolas: (ii) x2 = 4x + 4y

27. Find the equation of tangent to the parabola y2 = 2x + (i) y2 = x + 2
5y – 8 at x = 1. (iii) y2 = 4x + 4y – 8

28. Find the equation of the tangent to the parabola y2 = 8x NORMAL AND NORMALCY
having slope 2 and also find its point of contact.
50. Find the point of intersection of normals at P(t1) and
29. Two tangents are drawn from a point (–1, 2) to a pa- Q(t2) on the parabola y2 = 4ax.
rabola y2 = 4x. Find the angle between the tangents.
51 Find the relation between t1 and t2, where the normal at
30. Find the equation of the tangents to the parabola t1 to the parabola y2 = 4ax meets the parabola y2 = 4ax
y = x2 – 3x + 2 from the point (1, –1). again at t2.

31. Find the equation of the common tangent to the 52. If the normal at t1 meets the parabola again at t2, prove
parabolas y2 = 4ax and x2 = 4ay. that the minimum value of t22 is 8.

32. Find the equation of the common tangent to the 53. If two normals at t1 and t2 meet again the parabola y2 =
parabola y2 = 4ax and x2 = 4by. 4ax at t3, prove that t1t2 = 2.

33. Find the equation of the common tangent to the 54. Find the equation of the normal to the parabola y2 = 4x
parabola y2 = 16x and the circle x2 + y2 = 8.
at the point (1, 2).
34. Find the equation of the common tangents to the 55. Find the equation of the normal to the parabola y2 = 8x
parabolas y = x2 and y = –(x – 2)2.
at m = 2.
35. Find the equation of the common tangents to the curves 56. If x + y = k is a normal to the parabola y2 = 12x, find the
y2 = 8x and xy = –1.
value of k.
36. Find the equation of the common tangent to the circle 57. If the normal at P(18, 12) to the parabola y2 = 8x cuts it
x2 + y2 – 6y + 4 = 0 and the parabola y2 = x.
again at Q, prove that 9PQ = 80 10 .
37. Find the equation of the common tangent touching the
circle x2 + (y – 3)2 = 9 and the parabola y2 = 4x above 58. Find the locus of the point of intersection of two nor-
the x-axis. mals to the parabola y2 = 4ax, which are at right angles

38. Find the points of intersection of the tangents at the to one another.
ends of the latus rectum to the parabola y2 = 4x. 59. If lx + my + n = 0 is a normal to the parabola y2 = 4ax,

39. Find the angle between the tangents drawn from a point prove that al3 + 2alm2 + m2n = 0.
(1, 4) to the parabola y2 = 4x.
60. If a normal chord subtends a right angle at the vertex
40. Find the shortest distance between the line y = x – 2 and of the parabola y2 = 4ax, prove that it is inclined at an
the parabola y = x2 + 3x + 2. angle of tan-1( 2) to the axis of the parabola.

41. Find the shortest distance from the line x + y = 4 and the 61. At what point on the parabola y2 = 4x, the normal
parabola y2 + 4x + 4y = 0.
makes equal angles with the axes?
42. If y + b = m1(x + a) and y + b = m2(x + a) are two tan-
gents of the parabola y2 = 4ax, find the value of m1m2. 62. Find the length of the normal chord which subtends an
angle of 90° at the vertex of the parabola y2 = 4x.
43. The tangent to the curve y = x2 + 6 at a point (1, 7)
touches the circle x2 + y2 + 16x + 12y + c = 0 at Q. Find 63. Prove that the normal chord of a parabola y2 = 4ax at
the co-ordinates of Q.
the point (p, p) subtends a right angle at the focus.
44. Two straight lines are perpendicular to each other. One
of them touches the parabola y2 = 4a(x + a) and the 64. Show that the locus of the mid-point of the portion of
other touches y2 = 4b(x + b). Prove that the point of the normal to the parabola y2 = 4ax intercepted between
intersection of the lines lie on the line x + a + b = 0.
the curve and the axis is another parabola.
45. Prove that the area of the triangle formed by three 65. Find the shortest distance between the curves y2 = 4x

points on a parabola is twice the area of the triangle and x2 + y2 – 12x + 31 = 0.
66. Find the shortest distance between the curves x2 + y2 +
formed by the tangents at these points.
12y + 35 = 0 and y2 = 8x.
46. Prove that the circle circumscribing the triangle formed
CO-NORMAL POINT
by any three tangents to a parabola passes through the
67. Prove that the algebraic sum of the three concurrent
focus. normals to a parabola is zero.

68. Prove that the algebraic sum of the ordinates of the feet
of three normals drawn to a parabola from a given point
is also zero.

Parabola 4.11

69. Prove that the centroid of the triangle formed by the 85. Prove that the locus of the mid-points of the focal chord
of the parabola is another parabola.
feet of the three normals lies on the axis of the parabo-
86. Prove that the locus of the mid-points of the chord of a
la. Also find the centroid of the triangle. parabola passes through the vertex is a parabola.

70. If three normals drawn from a given point (h, k) to any 87. Prove that the locus of the mid-points of a normal
chords of the parabola y2 = 4ax is y4 – 2a(x – 2a)y2 +
parabola be real, prove that h > 2a. 8a4 = 0

71. If three normals from a given point (h, k) to any pa- 88. Prove that the locus of the mid-point of a chord of a
rabola y2 = 4ax be real and distinct, prove that 27ak2 < parabola y2 = 4ax which subtends a right angle at the
4(h – 2a)3. vertex is y2 = 2a(x – 4a).

72. If a normal to a parabola y2 = 4ax makes an angle q 89. Prove that the locus of the mid-points of chords of the
parabola y2 = 4ax which touches the parabola y2 = 4bx
with the axis of the parabola, prove that it will cut the is y2(2a – b) = 4a2x.

curve again at an angle of tan -1 Ê tan q ˆ . 90. Find the locus of the mid-point of the chord of the pa-
ËÁ 2 ˜¯ rabola y2 = 4ax, which passes through the point (3b, b).

73. Prove that the normal chord to a parabola y2 = 4ax at 91. Prove that the locus of the mid-points of all tangents
drawn from points on the directrix to the parabola
the point whose ordinate is equal to its abscissa, which y2 = 4ax is y2(2x + a) = a(3x + a)2.

subtends a right angle at the focus of the parabola. DIAMETER OF A PARABOLA

74. Prove that the normals at the end-points of the latus 92. Prove that the tangent at the extremity of a diameter of
rectum of a parabola y2 = 4ax intersect at right angle on a parabola is parallel to the system of chords it bisects.

the axis of the parabola and their point of intersection 93. Prove that tangents at the end of any chord meet on the
diameter which bisects the chords.
is (3a, 0).
REFLECTION PROPERTY OF A PARABOLA
75. If S be the focus of the parabola and the tangent and the
94. A ray of light moving parallel to the x-axis gets reflect-
normal at any point P meet the axes in T and G respec- ed from a parabolic mirror whose equation is (y – 4)2
= 8(x + 1). After reflection, the ray passes through the
tively, prove that ST = SG = SP. point (a, b), find the value of a + b + 10.
76. From any point P on the parabola y2 = 4ax, a perpen-
95. A ray of light is moving along the line y = x +2, gets
dicular PN is drawn on the axis meeting at N, the nor- reflected from a parabolic mirror whose equation is
y2 = 4(x + 2). After reflection, the ray does not pass
mal at P meets the axis in G. Prove that the sub-normal through the focus of the parabola. Find the equation of
the line which containing the reflected ray.
NG is equal to its semi-latus rectum.
77. The normal to the parabola y2 = 4ax at a point P on it,

meets the x-axis in G, prove that P and G are equidis-

tant from the focus S of the parabola.
78. The normal at P to the parabola y2 = 4ax meets its axis

at G. Q is another point on the parabola such that QG

is perpendicular to the axis of the parabola. Prove that
QG2 – PG2 = constant.

CHORD OF CONTACT LEVEL II
(Mixed Problems)
79. Find the equation of the chord of contact to the tangents
from the point (2, 3) to the parabola y2 = 4x. 1. Three normals to the parabola y2 = x are drawn through
a point (c, 0), then
80. Find the chord of contact of the tangents to the parabola (a) c = 1/4 (b) c = 1/2 (c) c > 1/2 (d) none
y2 = 12x drawn through the point (–1, 2).
2. The line which is parallel to x-axis and crosses the
81. Prove that the locus of the point of intersection of two
tangents to a parabola y2 = 4ax which make a given curve y = x at an angle of 45° is
angle q with one another is y2 – 4ax = (x + a)2 tan2q.
(a) x = 1/4 (b) y = 1/4 (c) y = 1/2 (d) y = 1
82. Prove that the length of the chord of contact of tan-
gents drawn from (h, k) to the parabola y2 = 4ax is 3. Consider a circle with its centre lying on the focus of
1 |(k 2 + 4a2 )(k 2 - 4ah)|1/2 . the parabola y2 = 2px such that it touches the directrix
a
of the parabola, the point of intersection of the circle
83. Prove that the area of the triangle formed by the tan-
gents from the point (h, k) to the parabola y2 = 4ax and and the parabola is
a chord of contact is (k 2 - 4ah)3/2 .
2a (a) ÊÁË p, p˜¯ˆ (b) Ê p , - pˆ¯˜
2 ÁË 2
CHORD BISECTED AT A POINT
(c) ÁËÊ - p, p˜ˆ¯ (d) ÁÊË - p , - pˆ˜¯
84. Find the equation of the chord of the parabola y2 = 8x 2 2
which is bisected at (2, 3).

4.12 Coordinate Geometry Booster

4. If the line x – 1 = 0 is the directrix of the parabola 15. If a π 0 and the line 2bx + 3cy + 4d = 0 passes through
y2 – kx + 8 = 0, one of the value of k is the point of intersection of the parabolas y2 = 4ax and
x2 = 4ay, then
(a) 1/8 (b) 8 (c) 4 (d) 1/4
(a) d2 + (2b – 3c)2 = 0 (b) d2 + (2b + 3c)2 = 0
5. If x + y = k is a normal to the parabola y2 = 12x, the (c) d2 + (3b + 2c)2 = 0 (d) d2 + (3b – 2c)2 = 0

value of k is

(a) 9 (b) 3 (c) –9 (d) –3 16. The locus of the vertices of the family of parabolas
y = a2x2 + a2x - 2a is
6. The equation of the directrix of the parabola y2 + 4y + 22

4x + 2 = 0 is

(a) x = –1 (b) x = 1

(c) x = –3/2 (d) x = 3/2 (a) xy = 105 (b) xy = 3
64 4
7. The equation of the common tangent touching the
circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above (c) xy = 35 (d) xy = 64
16 105
the x-axis is

(a) 3y = 3x + 1 (b) 3y = - (x + 3) 17. The angle between the tangents to the curve y = x2 – 5x
+ 6 at the point (2, 0) and (3, 0) is
(c) 3y = (x + 3) (d) 3y = - (3x + 1)

8. The locus of the mid-point of the line segment joining (a) p (b) p (c) p (d) p
the focus to a moving point on the parabola y2 = 4ax is 2 3 6 4

another parabola with directrix 18. The equation of a tangent to the parabola y2 = 8x is

(a) x + a = 0 (b) x = –a/2 y = x + 2. The point on this from which the other tangent

(c) x = 0 (d) x = a/2 to the parabola is perpendicular to the given tangent is

9. The focal chord of y2 = 16x is a tangent to (x – 6)2 + y2 (a) (–2, 0) (b) (–1, 1) (c) (0, 2) (d) (2, 4)

= 2, the possible values of the slope of this chord are 19. A parabola has the origin as its focus and the line x = 2

(a) 1, –1 (b) –1/2, 2 as its directrix. The vertex of the parabola is at

(c) –2, 1/2 (d) 1/2, 2 (a) (0, 1) (b) (2, 0)

10. The tangent to the parabola y = x2 + 6 touches the circle (c) (0, 2) (d) (1, 0)

x2 + y2 + 16x + 12y + c = 0 at the point 20. The length of the chord of the parabola y2 = x which is

(a) (–6, –9) (b) (–13, –9) bisected at the point (2, 1) is

(c) (–6, –7) (d) (13, 7) (a) 2 3 (b) 4 3 (c) 3 2 (d) 2 5

11. The axis of a parabola is along the line y = x and the 21. Two mutually perpendicular tangents to the parabola

distance of its vertex from the origin is 2 and that y2 = 4ax meet the axis in P1 and P2. If S be the focus of
the parabola, 1 + 1 is
from its focus is 2 2 . If the vertex and the focus both
l(SP1) l(SP2 )
lie in the first quadrant, then the equation of the pa-
(a) 4 (b) 2 (c) 1 (d) 1
rabola is a a a 4a

(a) (x + y)2 = (x + y – 2) (b) (x – y)2 = (x + y – 2) 22. Which of the following equations represents a para-
(c) (x – y)2 = 4(x + y – 2) (d) (x – y)2 = 8(x + y – 2) bolic profile, represented parametrically, is

12. The equations of the common tangents to the parabola (a) x = 3 cos t, y = 4 sin t
y = x2 and y = –(x – 2)2 is

(a) y = 4(x – 1) (b) y = 0 (b) x2 - 2 = -2 cos t, y = 4 cos2 Ê tˆ
ËÁ 2˜¯
(c) y = –4(x – 1) (d) y = –10(3x + 5)

13. The tangent PT and the normal PN to the parabola (c) x = tan t, y = sec t
y2 = 4ax at a point P on it meet its axis at points T and
Ê tˆ Ê tˆ
N, respectively. The locus of the centroid of the triangle (d) x= 1- sin t, y = sin ËÁ 2¯˜ + cos ÁË 2˜¯

PTN is a parabola whose

(a) vertex is Ê 2a , 0ˆ¯˜ (b) directrix is x = 0 23. The points of contact Q and R of tangent from the point
ËÁ 3 P(2, 3) on the parabola y2 = 4x are

(c) latus rectum is 2a (d) focus is (a, 0) (a) (9, 6), (1, 2) (b) (1, 2), (4, 4)

14. The normal at the point (bt12, 2bt1) on a parabola meets (c) (4, 4), (9, 6) (d) (9, 6), Ê 1 , 1˜¯ˆ
the parabola again at (bt22, 2bt2), then ÁË 4

(a) t2 = - t1 + 2 (b) t2 = t1 - 2 24. A tangent is drawn to the parabola y2 = 4x at the point
t1 t1
P whose abscissa lies in [1, 4]. The maximum possible

2 2 area of the triangle formed by the tangents at P ordinate
t1 t1
(c) t2 = t1 + (d) t2 = - t1 - of the point P and the x-axis is

(a) 8 (b) 16 (c) 24 (d) 32

Parabola 4.13

25. The length of the normal chord y2 = 4x which makes an and OQ as diameters intersect in R. If q1, q2 and j are
the angles made with the axis by the tangents at P and
angle of p with the axis of x is
4 Q on the parabola and by OR, the value of cot (q1) + cot

(a) 8 (b) 8 2 (c) 4 (d) 4 2 (q2) is

26. The co-ordinates of the end points of a focal chord (a) –2 tan (j) (b) –2 tan (p – j)

of a parabola y2 = 4ax are (x1, y1) and (x2, y2), then (c) 0 (d) –2 cot (j)

(x1x2 + y1y2) is 36. The tangent at P to a parabola y2 = 4ax meets the
(a) 2a2 (b) –3a2
(c) –a2 (d) 4a2 directrix at U and the base of the latus rectum at V, then

27. If the normal to a parabola y2 = 4ax at P meets the curve SUV (where S is the focus) must be a/an

again at Q and if PQ and the normal at Q makes angles (a) right D (b) equilateral D

a and b, respectively with the x-axis, then tan a + tan b (c) isosceles D (d) right isosceles D

is 37. Two parabolas y2 = 4a(x – m1) and x2 = 4a(y – m2)
always touch one another, the quantities m1 and m2 are
(a) 0 (b) –2 (c) –1/2 (d) –1 both variables. The locus of their points of contact has

28. If the normal to the parabola y2 = 4ax at the point with

parameter t1 cuts the parabola again at the point with the equation

parameter t2, then (b) 2 £ t22 £ 4 (a) xy = a2 (b) xy = 2a2
(a) 2 £ t22 £ 8
(c) t22 ≥ 4 (d) t22 ≥ 8 (c) xy = 4a2 (d) none

29. A parabola y = ax2 + bx + c crosses the x-axis at (a, 0) 38. If a normal to a parabola y2 = 4ax makes an angle

and (b, 0) both to the right of the origin. j with its axis, it will cut the curve again at an angle

A circle also passes through these two points. The (a) tan–1(2 tan j) (b) tan —1 Ê 1 tan j ˆ
length of a tangent from the origin to the circle is ËÁ 2 ¯˜

bc (c) b c (c) cot —1 Ê 1 tan j ˆ (d) none
(a) a a (d) a ËÁ 2 ˜¯
(b) ac2

30. Two parabolas have the same focus. If their directrices 39. The vertex of a parabola is (2, 2) and the co-ordinates

are the x-axis and the y-axis respectively, the slope of of its extremities of the latus rectum are (–2, 0) and

their common chord is (6, 0). The equation of the parabola is
(a) y2 – 4y + 8x – 12 = 0 (b) x2 + 4x + 8y – 12 = 0
(a) 1, –1 (b) 4/3 (c) 3/4 (d) none (c) x2 – 4x + 8y – 12 = 0 (d) x2 + 4x – 8y + 20 = 0
40. The length of the chord of the parabola y2 = x which is
31. The straight line joining the point P on the parabola bisected at the point (2, 1) is
y2 = 4ax to the vertex and the perpendicular from the

focus to the tangent at P intersect at R. Then the locus

of R is (a) 2 3 (b) 4 3 (c) 3 2 (d) 2 5

(a) x2 + 2y2 – ax = 0 (b) x2 + y2 – 2ax = 0 41. If the tangent and the normal at the extremities of a

(c) 2x2 + 2y2 – ax = 0 (d) 2x2 + y2 – 2ay = 0 focal chord of a parabola intersect at (x1, y1) and (x2, y2)
respectively, then
32. A normal chord of the parabola y2 = 4x subtending a

right angle at the vertex makes an acute angle q with (a) x1 = x2 (b) x1 = x2

the x-axis, then q is (c) y1 = y2 (d) x2 = y1

(a) tan–12 (b) sec-1( 3) 42. If the chord of contact of tangents from a point P to the

(c) cot-1( 3) (d) none parabola y2 = 4ax touches the parabola xy = 4by, then

33. C is the centre of the circle with centre (0, 1) and the the locus of P is a/an
radius unity of the parabola y = ax2. The set of values of
(a) circle (b) parabola

a for which they meet at a point other than the origin is (c) ellipse (d) hyperbola

(b) 0 < a < 1 43. The latus rectum of a parabola whose focal chord PSQ
2
(a) a > 0 is such that SP = 3 and SP = 2 is given by
(c) 1 < a < 1 (d) a > 1
2 (a) 24/5 (b) 12/5 (c) 6/5 (d) none
42
44. If two normals to a parabola y2 = 4ax intersect at right

34. TP and TQ are two tangents to the parabola y2 = 4ax at angles, then the chord joining then feet passes through

P and Q. If the chord PQ passes through the fixed point a fixed point whose co-ordinates are

(–a, b), the locus of T is (a) (–2a, 0) (b) (a, 0) (c) (2a, 0) (d) None

(a) ay = 2b(x – b) (b) by = 2a(x – a) 45. The straight line passing through (3, 0) and cutting the

(c) by = 2a(x – a) (d) ax = 2a(y – b) curve y = x orthogonally is

35. Through the vertex O of the parabola y2 = 4ax two (a) 4x + y = 18 (b) x + y = 9

chords OP and OQ are drawn and the circles on OP (c) 4x – y = 6 (d) None

4.14 Coordinate Geometry Booster

46. PQ is a normal chord of the parabola y2 = 4ax at P. A 8. If the focal chord of y2 = 16x is a tangent to (x – 6)2 + y2

being the vertex of the parabola. Through P a line is = 2, find the possible values of the slope of the chord.

drawn parallel to AQ meeting the x-axis in R. Then the 9. Find the points of intersection of the tangents at the end
of the latus rectum of the parabola y2 = 4x.
length of AR is
10. If the tangent at the point P(2, 4) to the parabola
(a) the length of latus rectum. y2 = 8x meets the parabola y2 = 8x + 5 at Q and R, find

(b) the focal distance of the point P. the mid-point of QR.

(c) 2 ¥ Focal distance of the point P. 11. If y + b = m1(x + a) and y + b = m2(x + a) are two tan-
gents to the parabola y2 = 4ax, prove that m1m2 = –1.
(d) the distance of P from the directrix.
12. Find the equation of the common tangent to the circle
47. The locus of the point of intersection of the perpen- (x – 3)2 + y2 = 9 and the parabola y2 = 4x above the
dicular tangents of the curve y2 + 4y – 6x – 2 = 0 is
x-axis.
(a) 2x = 1 (b) 2x + 3 = 0
13. Find the equation of the common tangent to the curves
(c) 2y + 3 = 0 (d) 2x + 5 = 0 y2 = 8x and xy = –1.

48. The length of the focal chord of the parabola y2 = 4ax 14. Find the common tangents of y = x2 and y = –x2 + 4x – 4.

at a distance p from the vertex is 15. Consider a circle with the centre lying on the focus of
the parabola y2 = 2px such that it touches the direc-
2a2 a3 4a3 p3
(a) (b) p2 (c) p2 (d) trix of that parabola. Find a point of intersection of the

p a circle and the parabola.
16. If the normal drawn at a point (at12, 2at1) of the
49. The locus of a point such that two tangents drawn from
it to the parabola y2 = 4ax are such that the slope of one parabola y2 = 4ax meets it again at (at22, 2at2), prove
that t12 + t1t2 + 2 = 0.
is double the other is 17. Three normals to the parabola y2 = x are drawn through

(a) y2 = 9 ax (b) y2 = 9 ax a point (c, 0), find c.
2 4 18. A tangent to the parabola y2 = 8x makes an angle 45°

(c) y2 = 9ax (d) x2 = 4ay with the straight line y = 3x + 5. Find the equation of

50. The point on the parabola y2 = 4x which are closest to the tangent and its point of contact.
the curve x2 + y2 – 24y + 128 = 0 is 19. Find the equations of the normal to the parabola y2 =

(a) (0, 0) (b) (2, 2) 4ax at the ends of the latus rectum. If the normal again
meets the parabola at Q and Q¢, prove that QQ¢ = 12a.
(c) (4, 4) (d) none 20. Prove that from any point P(at2, 2at) on the parabola
y2 = 4ax, two normals can be drawn and their feet
LEVEL III
(Problems for JEE Advanced) Q and R have the parameters satisfying the equation
l2 + lt + 2 = 0.
1. If two ends of a latus rectum of a parabola are the points
(3, 6) and (–5, 6), find its focus. 21. Find the locus of the points of intersection of those nor-
mals to the parabola x2 = 8y which are at right angles to
2. Find the locus of the focus of the family of parabolas
y = a2x2 + a2x - 2a . each other.
32
22. Two lines are drawn at right angles, one being a tangent
3. Find the equation of the parabola whose vertex and the to y2 = 4ax and the other to x2 = 4by. Show that the
focus lie on the axis of x at distances a and a1 from the
origin, respectively. locus of their points of intersection is the curve
(ax + by)(x2 + y2) + (bx – ay)2 = 0.
4. A square has one vertex at the vertex of the parabola
y2 = 4ax and the diagonal through the vertex lies 23. Prove that the locus of the centroid of an equilat-
along the axis of the parabola. If the ends of the other eral triangle inscribed in the parabola y2 = 4ax is
diagonal lie on the parabola, find the co-ordinates of 9y2 = 4a(x – 8a).
the vertices of the square.
24. Prove that the locus of the mid-points of chords of the
5. Find the equation of the parabola whose axis is y = x, parabola y2 = 4ax which subtends a right angle at the
vertex is y2 = 2a2(x – 4).
the distance from origin to vertex is 2 and the dis-
25. Prove that the locus of a point that divides a chord of
tance from the origin to the focus is 2 2 . slope 2 of the parabola y2 = 4x internally in the ratio
6. If a π 0 and the line 2bx + 3cy + 4d = 0 passes through
1 : 2 is a parabola. Find the vertex of the parabola.
the points of intersection of the parabola y2 = 4ax and
x2 = 4ay, prove that d2 + (2b + 3c)2 = 0. 26. From a point A, common tangents are drawn to the
7. Two straight lines are perpendicular to each other. One
of them touches the parabola y2 = 4a(x + a) and the circle x2 + y2 = a2 and the parabola y2 = 4ax. Find the
other y2 = 4b(x + b). Prove that their points of intersec- 2
tion lie on the line x + a + b = 0.
area of the quadrilateral formed by the common tan-

gents, the chord of contact of the circle and the chord

of contact of the parabola.

Parabola 4.15

27. Normals are drawn from the point P with slopes m1, 7. The ordinates of points P and Q on the parabola
m2, m3 to the parabola y2 = 4x. If the locus of P with y2 = 12x are in the ratio 1 : 2. Find the locus of the points
m1m2 = a is a part of the parabola, find the value of a.
of intersection of the normals to the parabola at P and Q.
28. The tangent at a point P to the parabola y2 – 2y – 4x +
[Roorkee Main, 1998]

5 = 0 intersects the directrix at Q. Find the locus of a 8. Find the equations of the common tangents of the circle

point R such that it divides PQ externally in the ratio x2 + y2 – 6y + 4 = 0 and y2 = x. [Roorkee, 1999]

1 :1. 9. Find the locus of points of intersection of tangents

2 drawn at the ends of all normals chords to the parabola
Prove that the curve
29. y = - x2 + x + 1 is symmetric y2 = 4(x – 1). [Roorkee Main, 2001]

2 10. Find the locus of the trisection point of any double
ordinate y2 = 4ax.
with respect to the line x = 1. And also prove that it is
11. Find the locus of the trisection point of any double
symmetric about its axis.
ordinate x2 = 4by.
30. Three normals are drawn from the point (14, 7) to the 12. Find the shortest distance between the curves y2 = 4x
parabola y2 – 16x – 8y = 0. Find the co-ordinates of the
and x2 + y2 – 12x + 31 = 0.
feet of the normals.
13. Find the radius of the circle that passes through the
31. Find the locus of the foot of the perpendicular drawn
origin and touches the parabola at (a, 2a).
from a fixed point to any tangent to a parabola. 14. Find the condition if two different tangents of y2 = 4x

32. Find the locus of the point of intersection of the are the normals to x2 = 4by.
normals to the parabola y2 = 4ax at the extremities of
15. A circle is drawn to pass through the extremities of the

a focal chord. latus rectum of the parabola y2 = 8x. It is also given that

the circle touches the directrix of the parabola. Find the

LEVEL IV radius of the circle.
(Tougher Problems for JEE
Advanced) Integer Type Questions

1. Find the maximum number of common chords of a

1. From the point (–1, 2), tangent lines are drawn to the parabola and a circle.
parabola y2 = 4x. Find the equation of the chord of con-
2. If the straight lines y – b = m1(x + a) and y – b =
tact. Also find the area of the triangle formed by the m2(x + a) are the tangents of y2 = 4ax, find the value of
(m1m2 + 4).
chord of contact and the tangents.
3. A normal chord of y2 = 4ax subtends an angle p at
[Roorkee Main, 1994] 2
2. The equation y2 – 2x – 2y + 5 = 0 represents
the vertex of the parabola. If its slope is m, then find the
(a) a circle with centre (1, 1)
value of (m2 + 3).
(b) a parabola with focus (1, 2) 4. Find the slope of the normal chord of y2 = 8x that gets

(c) a parabola with directrix x = 3/2 bisected at (8, 2).
(d) a parabola with directrix x = –1/2
5. Find the maximum number of common normals of y2 =
[Roorkee, 1995] 4ax and x2 = 4by.
3. A ray of light is coming along the line y = b from the
6. Find the length of the latus rectum of the parabola
positive direction of x-axis and strikes a concave mir-
ror whose intersection with the xy-plane is a parabola whose parametric equation are given by x = t2 + t + 1
y2 = 4ax. Find the equation of the reflected ray and and y = t2 – t + 1.
show that it passes through the focus of the parabola. 7. If the shortest distance between the curves y2 = x – 1
and x2 = y – 1 is d, find the value of (8d2 – 3).
Both a and b are positive. [Roorkee Main, 1995]
8. If m1 and m2 be the slopes of the tangents that are drawn
4. If a tangent drawn at a point (t2, t2) on the parabola from (2, 3) to the parabola y2 = 4x, find the value of
y2 = 4x is the same as the normal drawn at a point
Ê1 + 1 + 2ˆ¯˜ .
( 5 cos j, 2 sin j) on the ellipse 4x2 + 5y2 = 20, find ÁË m1 m2

the values of t and j. [Roorkee, 1996] 9. Let P(t1) and Q(t2) are two points on the parabola y2 =
4ax. If the normals at P and Q meet the parabola again
5. The equations of normals to the parabola y2 = 4ax at the

point (5a, 2a) are at R, find the value of (t1t2 + 3).

(a) y = x – 3a (b) y = x + 3a 10. Find the length of the chord intercepted between the

(c) y + 2x = 12a (d) 3x + y = 33a parabola y2 = 4x and the straight line x + y = 1.
11. If x + y = k is a normal to the parabola y2 = 12x, find the
[Roorkee, 1997]

6. Find the locus of the points of intersection of those nor- value of k.

mals to the parabola x2 = 8y which are at right angles to 12. Find the number of distinct normals drawn from the
point (–2, 1) to the parabola y2 – 4x – 2y – 3 = 0.
each other. [Roorkee Main, 1997]

4.16 Coordinate Geometry Booster

Comprehensive Link Passages (iii) The equation of the directrix of the parabola is

(a) 6x + 7 = 0 (b) 4x + 11 = 0

Passage I (c) 3x + 11 = 0 (d) 2x + 13 = 0
y = x is a tangent to the parabola y2 = ax2 + c.
Passage V
(i) If a = 2, the value of c is The normals at three points P, Q, R on the parabola y2 = 4ax

(a) 1 (b) –1/2 (c) 1/2 (d) 1/8 meet at (a, b).

(ii) If (1, 1) is a point of contact, the value of a is (i) The centroid of DPQR is

(a) 1/ 4 (b) 1/3 (c) 1/ 2 (d) 1/6 ÁËÊ a - 2a 0˜ˆ¯ ÊËÁ 2a - 4a 0¯ˆ˜
3 3
(iii) If c = 2, the point of contact is (a) , (b) ,

(a) (3, 3) (b) (2, 2) (c) (6, 6) (d) (4, 4). a b ËÁÊ a + b b - a
3 3 3 3
Passage II (c) ÁËÊ , ˜ˆ¯ (d) , ˜¯ˆ
Consider, the parabola whose focus is at (0, 0) and the tan-
gent at the vertex is x – y + 1 = 0. (ii) The orthocentre of DPQR is

(i) The length of the latus rectum is (a) ÊÁË a + 6a, b ˆ˜¯ (b) ÁÊË a + 3a, b ˜ˆ¯
2 2
(a) 4 2 (b) 2 2 (c) 8 2 (d) 3 2

(ii) The length of the chord of the parabola on the x-axis is (c) ÁÊË a - 6a, - b ¯ˆ˜ (d) ÁËÊ a - 3a, b ˆ˜¯
2 2

(a) 4 2 (b) 2 2 (c) 8 2 (d) 3 2 (iii) The circumcentre of DPQR is

(iii) Tangents drawn to the parabola at the extremities of the (a) ÊËÁ a + 2a , - b ¯˜ˆ (b) ËÁÊ a + 2a , b ˆ˜¯
2 4 2 4
chord 3x + 2y = 0 intersect at an angle

(a) p (b) p (c) p (d) none ÁÊË a b ˆ˜¯ ËÁÊ a b ¯ˆ˜
6 3 2 4 4 2 2
(c) , (d) ,

Passage III

Two tangents on a parabola are x – y = 0 and x + y = 0. If (2, 3) Passage VI
Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They
is the focus of the parabola. intersect at P and Q in the first and the fourth quadrants, re-
spectively. The tangents to the circle at P and Q intersect the
(i) The equation of the tangent at the vertex is x-axis at R and tangents to the parabola at P and Q intersect
the x-axis at S.
(a) 4x – 6y + 5 = 0 (b) 4x – 6y + 3 = 0
(i) The ratio of the areas of Ds PQS and PQR is
(c) 4x – 6y + 1 = 0 (d) 4x – 6y + 3/2 = 0
(a) 1: 2 (b) 1 : 2 (c) 1 : 4 (d) 1 : 8
(ii) The length of the latus rectum of the parabola is

(a) 6 (b) 10 (c) 2 (d) 9 2
3 13 13 13

(iii) If P and Q are ends of the focal chord of the parabola, (ii) The radius of the circumcircle of DPRS is
then is
(a) 5 (b) 3 3 (c) 3 2 (d) 2 3

2 13 2 13 2 13 (iii) The radius of the incircle of DPQR is
(a) 3 (b) 2 13 (c) 3 (d) 7
(a) 4 (b) 3 (c) 8/3 (d) 2

Passage IV Passage VII
If l, m are variable real numbers such that 5l2 + 6m2 – 4lm + 3l If P is a point moving on a parabola y2 = 4ax and Q is a mov-
ing point on the circle x2 + y2 – 24ay + 128a2 = 0. The points
= 0, the variable line lx + my = 1 always touches a fixed pa-
P and Q will be closest when they lie along the normal to the
rabola, whose axis is parallel to x-axis. parabola y2 = 4ax passing through the centre of the circle.

(i) The vertex of the parabola is (i) If the normal at (at2, 2at) of the parabola passes through

(a) ÊËÁ - 5, 4 ˜¯ˆ (b) ÊËÁ - 7 , 3 ¯˜ˆ the centre of the circle, the value of t must be
3 3 4 4
(a) 1 (b) 2 (c) 3 (d) 4

(c) ÊËÁ 5 , - 76 ˆ˜¯ (d) ÊËÁ 1 , - 3 ˆ¯˜ (ii) The shortest distance between P and Q must be
6 2 4
(a) a( 2 - 1) (b) 2a( 5 - 1)

(ii) The focus of the parabola is (c) 4a( 5 - 1) (d) 4a( 5 + 1)

(a) ËÁÊ 1 , - 76 ˜¯ˆ (b) ÊËÁ 1, 34¯ˆ˜ (iii) When P and Q are closest, the point P must be
6 3
(b) (2a, 2 2a)
(c) ËÊÁ 3 , - 3 ˆ¯˜ (d) ËÊÁ - 3 , 3 ˆ˜¯ (a) (1, 2a) (d) (5a, 4a)
2 2 4 4 (c) (4a, 4a)

Parabola 4.17

(iv) When P and Q are closest, the point Q must be 2. Match the following columns:

(a) Ê 4a , 6a - 8a ˆ (b) Ê 4a , 12a - 8a ˆ Column I Column II
ËÁ 5 5 ˜¯ ËÁ 5 5 ¯˜
(A) The point, from which perpen- (P) (–1, 2)

(c) Ê 4a , - 12a - 8a ˆ (d) Ê 4a , - 8a - 12aˆ dicular tangents can be drawn
ËÁ 5 5 ˜¯ ÁË 5 5 ˜¯ to the parabola y2 = 4x, is

(v) When P and Q are closest and the diameter QR of the (B) The point, from which only one (Q) (3, 2)

circle is drawn through Q, the x-cordinate of Q is normal can be drawn to the pa-
rabola y2 = 4x, is
(a) - 2a (b) - 8a (c) - 4a (d) - 6a
555 5 (C) The point, at which chord x – y (R) (–1, –5)
+ 1 = 0 of the parabola y2 = 4x
Passage VIII
is bisected, is

If a source of light is placed at the fixed point of a parabola (D) The point, from which tangents (S) (5, –2)

and if the parabola is reflecting surface, the ray will bounce cannot be drawn to the parabola
y2 = 4x, is
back in a line parallel to the axis of the parabola.

(i) A ray of light is coming along the line y = 2 from the 3. Match the following columns:

positive direction of x-axis and strikes a concave mirror Column I Column II
whose intersection with the xy plane is a parabola y2 =

8x, the equation of the reflected ray is (A) The equation of the director (P) 2y – 1 = 0
circle to the parabola y2 = 12x is
(a) 2x + 3y = 4 (b) 3x + 2y = 6

(c) 4x + 3y = 8 (d) 5x + 4y = 10 (B) The equation of the director (Q) x – 2 = 0
circle to the parabola x2 = 16y is
(ii) A ray of light moving parallel to the x-axis gets reflect-
ed from a parabolic mirror whose equation is y2 + 10y – (C) The equation of the director (R) y + 4 = 0

4x + 17 = 0. After reflection, the ray must pass through circle to the parabola
y2 + 4x + 4y = 0 is
the point

(a) (–2, –5) (b) (–1, –5)

(c) (–3, –5) (d) (–4, –5) (D) The equation of the director (S) x + 3 = 0

(iii) A ray of light is coming along the line x = 2 from the circle to the parabola
y = x2 + x + 1 is
positive direction of y-axis and strikes a concave mirror
whose intersection with the xy plane is a parabola x2 = 4. Match the following columns:

4y, the equation of the reflected ray after second reflec-

tion is Column I Column II

(a) 2x + y = 1 (b) 3x – 2y + 2 = 0 (A) Number of distinct normals can (P) 3

(c) y = 1 (d) none be drawn from (–2, 1) to the pa-
rabola y2 – 4x – 2y – 3 = 0 is
(iv) Two rays of light coming along the line y = 1 and

y = –2 from the positive direction of x-axis and strikes a (B) Number of distinct normals can (Q) 1

concave mirror whose intersection with the xy plane is be drawn from (2, 3) to the pa-
a parabola y2 = x at A and B, respectively. The reflected rabola y = x2 + x + 1 is

rays pass through a fixed point C, the area of DABC is (C) Number of distinct normals can (R) 0

(a) 21/ 8 s.u. (b) 19/2 s.u. be drawn from (–5, 3) to the pa-
rabola y2 – 4x – 6y – 1 = 0 is
(c) 17/2 s.u. (d) 15/2 s.u.

Matrix Match (D) Number of tangents can be drawn (S) 2
(For JEE-Advanced Examination Only)
from (1, 2) to the parabola
1. Match the following columns: AB is a chord of the pa- y2 – 2x – 2y + 1 = 0 is
rabola y2 = 4ax joining A(at12, 2at1) and B(at22, 2at2).
5. Match the following columns:
Column I Column II
Normals are drawn at points P, Q and R lying on the
(A) AB is a normal chord, if (P) t2 = 2 – t1 parabola y2 = 4x which intersect at (3, 0).

(B) AB is a focal chord, if (Q) t1t2 = –4 Column I Column II

(C) AB subtends 90° at (R) t1t2 = –1 (A) Area of DPQR (P) 2
(0, 0), if
(B) Radius of the circumcircle DPQR (Q) 5/2

(D) AB is inclined at 45° to (S) t12 + t1t2 + 2 = 0 (C) Centroid of DPQR (R) (5/2, 0)
the axis of the parabola
(D) Circumcentre of DPQR (S) (2/3, 0)

4.18 Coordinate Geometry Booster

6. Match the following columns: 2. A is a point on the parabola y2 = 4ax. The normal at
A cuts the parabola again at B. If AB subtends a right
Column I Column II angle at the vertex of the parabola, find the slope of AB.
[IIT-JEE -1982]
(A) If y + 3 = m1(x + 2) and (P) m1 + m2 = 0
y + 3 = m2(x + 2) are two No questions asked in 1983.
tangents to the parabola 3. Find the equation of the normal to the curve x2 = 4y
y2 = 8x, then
which passes through the point (1, 2). [IIT-JEE, 1984]
(B) If y = m1x + c1 and y = m2x (Q) m1 + m2 = –1 4. Three normals are drawn from the point (c, 0) to the
+ c2 are two tangents to a
parabola y2 = 4a(x + a), curve y2 = x. Show that c must be greater than 1/2. One
normal is always the x-axis. Find c for which the other
then two normals are perpendicular to each other.

(C) If y = m1(x + 4) + 2013 and (R) m1, m2 = –1 [IIT-JEE, 1991]
y + 1 = m2(x + 4) + 2014 5. Through the vertex O of the parabola y2 = 4x, chords
are two tangents to the
parabola y2 = 16x, then OP and OQ are drawn at right angles, show that for
all positions of P, PQ cuts the axis at the parabola at a
(D) If y = m1(x – 2) + 2010 (S) m1 – m2 = 0 fixed point. Also find the locus of the mid-point of PQ.
and y = m1(x – 2) + 2010
are two tangents to the [IIT-JEE, 1994]
parabola y2 = –8x, then 6. Consider a circle with its centre lying on the focus of

7. Match the following columns: the parabola y2 = 2px such that it touches the direc-
trix of the parabola. Then a point of intersection of the
circle and the parabola is

Column I Column II Ê p, p˜ˆ¯ Ê p pˆ¯˜ ËÁÊ p, p ˆ˜¯
ÁË 2 ËÁ 2 2 2
(A) If 2x + y + l = 0 is a normal to the (P) 9 (a) or - , (b)
parabola y2 = –8x, then l is

(B) If x + y = k is a normal to the pa- (Q) 24 (c) ËÊÁ - p, p¯˜ˆ (d) ÁËÊ - p , - p ¯˜ˆ
rabola y2 = 12x, then k is 2 2 2

(C) If 2x – y – c = 0 is a tangent to the (R) 17 [IIT-JEE, 1995]
parabola y2 = 16x, then c is 2
7. Show that the locus of a point that divides a chord of
(D) If y = 4x + d is a tangent to the (S)
parabola y2 = 4(x + 1), then 4d is slope 2 of the parabola y2 = 4x internally in the ratio

1 : 2 is parabola. Also find its vertex. [IIT-JEE, 1995]
8. Points A, B, C lie on the parabola y2 = 4ax. The tangent

8. Match the following columns to the parabola at A, B and C taken in pair intersect at

the points P, Q and R. Determine the ratio of the areas

Column I Column II of Ds ABC and PQR. [IIT-JEE, 1996]

(A) The equation of the directrix (P) x – 2y + 4 9. From a point A, common tangents are drawn to the
circle x2 + y2 = a2 and the parabola y2 = 4ax. Find
of the parabola =0 2

y2 + 4x + 4y + 2 = 0 is

(B) The equation of the axis of (Q) 2x – 3 = 0 the area of the quadrilateral formed by the common

the parabola tangents drawn from A and the chord of contact of the
x2 + 4x + 4y + 2013 = 0 is
circle and the parabola. [IIT-JEE, 1996]

(C) The equation of the tangent to (R) x + 3 = 0 No questions asked in between 1997 to 1999.
the parabola y2 = 4x from the
10. If x + y = k is normal to y2 = 12x, then k is
point (2, 3) is
(a) 3 (b) 9 (c) –9 (d) –3

(D) The equation of the directrix (S) x + 2 = 0 [IIT-JEE, 2000]
of the parabola y2 = 4x + 8 is 11. If the line x – 1 = 0 is the directrix of the parabola y2 =

kx – 8, one of the value of the k is

(a) 1/8 (b) 8 (c) 4 (d) 1/4

Questions asked in Previous Years’ [IIT-JEE, 2000]
JEE-Advanced Examinations 12. The equation of the directrix of the parabola y2 + 4y +

1. Suppose that the normals drawn at three different 4x + 2 = 0 is

points on the parabola y2 = 4x pass through the point (a) x = –1 (b) x = 1

(h, k). Show that h > 2a. [IIT-JEE, 1981] (c) x = –3/2 (d) x = 3/2

[IIT-JEE, 2001]

Parabola 4.19

13. The equation of the common tangent touching the 22. The equation of the common tangent to the parabola
circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above y = x2 and y = – (x – 2)2 is/are

the x-axis is (a) y = 4(x – 1) (b) y = 0

(a) y 3 = (3x + 1) (b) y 3 = - x - 3 (c) y = –4(x – 1) (d) y = –30x – 50

(c) y 3 = x + 3 (d) y 3 = - (3x + 1) [IIT-JEE, 2006]

26. Match the following:

[IIT-JEE, 2001] Normals are drawn at points P, Q and R lying on the
parabola y2 = 4x which intersect at (3, 0).
14. The locus of the mid-point of the line segment joining
the focus to a moving a point on the parabola y2 = 4x is Column I Column II
(i) Area of DPQR (A) 2
another parabola with directrix (ii) Radius of circumcircle of (B) 5/2
DPQR
(a) x = –a (b) x = –a/2 (C) (1, 0)
(iii) Centroid of DPQR (D) (2/3, 0)
(c) x = 0 (d) x = a/2 (iv) Circumcentre of DPQR

[IIT-JEE, 2002]
15. The equation of the common tangent to the curve y2 =

8x and xy = –1 is

(a) 3y = 9x + 2 (b) y = 2x + 1

(c) 2y = x + 8 (d) y = x + 2 [IIT-JEE, 2006]
24. Comprehension
[IIT-JEE, 2002]
16. The focal chord to y2 = 16x is tangent to (x – 6)2 + y2 = 2, Consider the circle x2 + y2 = 9 and the parabola y2 =
8x. They intersect at P and Q in the first and the fourth
the possible values of the slope of this chord, are quadrants, respectively. The tangents to the circle at P
and Q intersect the x-axis at R and tangents to the pa-
(a) {–1, 1} (b) {–2, 2} rabola at P and Q intersect the x-axis at S.
(i) The ratio of the areas of Ds PQS and PQR is
(c) {–2, 1/2} (d) {2, –1/2}

[IIT-JEE, 2003]

17. Let C1 and C2 be, respectively, the parabolas x2 = y – 1 (a) 1: 2 (b) 1 : 2 (c) 1 : 4 (d) 1 : 8
and y2 = x – 1, let P be any point on C1 and Q be any (ii) The radius of the circumcircle of DPRS is
point on C2. Let P1 and Q1 be the reflections of P and
Q respectively, with respect to the line y = x. Prove that (a) 5 (b) 3 3 (c) 3 2 (d) 2 3

P1 lies on C2, P1 lies on C1 and PQ ≥ min[PP1, QQ1]. (iii) The radius of the incircle of DPQR is

Hence or otherwise determine points P0 and Q0 on the (a) 4 (b) 3 (c) 8/3 (d) 2

parabolas P0 and Q0, C1 and C2, respectively such that y = - x2 + x [IIT-JEE, 2007]
P0Q0 £ PQ for all pairs of points (P1, Q) with P on C1 + 1 is symmetric

and Q on C2 [IIT-JEE, 2003] 25. Statement 1: The curve
with respect to the line x = 1. 2
18. Three normals with slopes m1, m2 and m3 are drawn
from a point P not on the axis of the parabola y2 = 4x.
Statement 2: A parabola is symmetric about its axis.

If m1m2 = a, results in the locus of P being a part of the [IIT-JEE, 2007]

parabola, find the value of a. [IIT-JEE, 2003] 26. Let P(x1, y1), Q(x2, y2), y1 < 0, y2 < 0, be the end-points
of the latus rectum of the ellipse x2 + 4y2 = 4. The equa-
19. The angle between the tangents drawn from the point

(1, 4) to the parabola y2 = 4x is tion of the parabola with the latus rectum PQ are

(a) p (b) p (c) p (d) p (a) x2 + 2 3y = 3 + 3 (b) x2 - 2 3y = 3 + 3
6 4 3 2 (c) x2 + 2 3y = 3 - 3 (d) x2 - 2 3y = 3 - 3

[IIT-JEE, 2004] [IIT-JEE, 2008]
20. At any point P on the parabola y2 – 2y – 4x + 5 = 0, a
27. The tangent PT and the normal PN to the parabola
tangent is drawn which meets the directrix at Q. Find
y2 = 4ax at a point P on it meet axes at points T and N,
the locus of point R which divides QP externally in the
respectively. The locus of the centroid of DPTN is a
ratio 1 :1.
2 [IIT-JEE, 2004] parabola whose

(a) vertex is (2a/3, 0) (b) directrix is x = 0

No questions asked in 2005. (c) latus rectum is 2a/3 (d) focus is (a, 0)

21. The axis of a parabola is along the line y = x and the [IIT-JEE, 2009]
distances of its vertex and focus from the origin are 28. Let A and B be two distinct points on the parabola y2 =

2 and 2 2 , respectively. If the vertex and the focus 4x. If the axis of the parabola touches a circle of radius
both lie in the first quadrant, the equation of the pa-
rabola is r having AB as the diameter, the slope of the line join-
(a) (x – y)2 = (x – y – 2) (b) (x – y)2 = (x + y – 2)
(c) (x – y)2 = 4(x + y – 2) (d) (x – y)2 = 8(x + y – 2) ing A and B can be

[IIT-JEE, 2006] (a) - 1 (b) 1 (c) 2 (d) - 2
r r r r

[IIT-JEE, 2010]

4.20 Coordinate Geometry Booster

29. Let L be a normal to the parabola y2 = 4x. If L passes 36. The value of r is

through the point (9, 6), then L is given by (a) - 1 (b) t2 + 1 (c) 1 (d) t2 - 1
t tt t
(a) y – x + 3 = 0 (b) y + 3x – 33 = 0

(c) y + x – 15 = 0 (d) y – 2x + 12 = 0 [IIT-JEE, 2104]

[IIT-JEE, 2011] 37. If st = 1, the tangent at P and the normal at S to the
30. Consider the parabola y2 = 8x, Let D1 be the area of the
parabola meet at a point whose ordinate is
triangle formed by the end-points of its latus rectum

and the point P ÁËÊ 1 , 2˜¯ˆ on the parabola, and D2 be the (a) (t1 + 1)2 a(t2 + 1)2
2 2t 3 (b) 2t3

area of the triangle formed by drawing tangents at P (c) a(t2 + 1)2 (d) a(t2 + 2)2
t3 t3
and at the end-points of the latus rectum. Then D1 is...
D2 [IIT-JEE, 2014]
38. Let the curve C be the mirror image of the parabola y2
[IIT-JEE, 2011]
31. Let S be the focus of the parabola y2 = 8x and let PQ be = 4x with respect to the line x + y + 4 = 0. If A and B are

the common chord of the circle x2 + y2 – 2x – 4y = 0 and

the given parabola. The area of DPQS is ... the points of intersection of C with the line y = –5, then

[IIT-JEE, 2012] the distance between A and B is... [IIT-JEE-2015]
39. If the normals of the parabola y2 = 4x drawn at the end
Comprehension
Let PQ be a focal chord of a parabola y2 = 4ax. The tangents points of its latus rectum are tangents to the circle (x –
3)2 + (y + 2)2 = r2, then the value of r2 is...
to the parabola at P and Q meet at a point lying on the line

y = 2x + a, a > 0. [IIT-JEE-2015]
40. Let P and Q be distinct points on the parabola y2 = 2x
32. The length of the chord PQ is
such that a circle with PQ as diameter passes through
(a) 7a (b) 5a (c) 2a (d) 3a
the vertex O of the parabola. If P lies in the first quad-
33. If the chord PQ subtends an angle q at the vertex of y2 rant and the area of the triangle DOPQ is 3: 2, then

= 4ax, then tan q is

(a) 2 7 (b) - 2 7 (c) 2 5 (d) - 2 5 which of the following is (are) the coordinates of P?
3 33 3
(a) (4, 2 2) (b) (9, 3 2)

[IIT-JEE, 2013] Ê 1 , 1ˆ (d) (1, 2)
ËÁ 4 2 ˜¯ [IIT-JEE-2015]
37. Match Matrix (c)

A line L: y = mx + 3 meets y-axis at E(0, 3) and the arc 41. The circle C1 : x2 + y2 = 3, with centre at O, intersects
of the parabola y2 = 16x, 0 £ y £ 6 at the point F(x0, y0). the parabola x2 = 2y at the point P in the first quadrant.

The tangent to the parabola at F(x0, y0) intersects the Let the tangent to the circle C1 at P touches other two
circles C2 and C3 at R2 and R3, respectively.
y-axis at G(0, y1). The slope m of the line L is chosen
such that the area of DEFG has a local maximum.

Match List I and List II and select the correct answers Suppose C2 and C3 have equal radii 2 3 and centres
Q2 and Q3, respectively.
using the code given below the lists. If Q2 and Q3 lie on the y-axis, then

List I List II

P m= 1. 1/2

Q Maximum area of DEFG is 2. 4 (a) Q2Q3 = 12 (b) R2R3 = 4 6

R y0 = 3. 2 (c) ar(DOR2R3) = 6 2 (d) ar(DPQ2Q3) = 4 2

S y1 = 4. 1 [IIT-JEE-2016]
42. Let P be the point on the parabola y2 = 4x which is at
[IIT-JEE, 2013]
the shortest distance from the center S of the circle x2 +
35. The common tangents to the circle x2 + y2 = 2 and the y2 – 4x – 16y + 64 = 0. Let Q be the point on the circle

parabola y2 = 8x touch the circle at the points P, Q and dividing the line segment SP internally. Then

the parabola at the points R, S. Then the area of the

quadrilateral PQRS is

(a) 3 (b) 6 (c) 9 (d) 15 (a) SP = 2 5

[IIT-JEE, 2014] (b) SQ:SP = ( 5 + 1):2

Comprehension (c) the x-intercept of the normal to the parabola at P is

Let a, r, s, t be non-zero real numbers. (d) 6. 1
Let P(at2, 2at), Q, R(ar2, 2ar) and S(as2, 2as) be distinct the slope of the tangent to the circle at Q is
points on the parabola y2 = 4ax. Suppose that PQ is the focal 2

chord and lines QR and PK are parallel, where K is the point [IIT-JEE-2016]

(2a, 0).

Parabola 4.21

ANSWERS

LEVEL I 33. y = ±x ± 4
34. y = 4x – 4
1. Parabola 35. y = x + 2
2. l = 4 36. x + 2y = 1
3. 1
37. x - 3y + 3 = 0
2
4. 20{x2 + y2 – 2x – 4y + 5} 38. L(1, 2) and L¢(1, –2) is (1, 0)
39. p
= (x2 + 9y2 + 100 + 6xy + 20x + 6y)
3
5. x2 + 2xy + y2 + 2x + 2y + 4 = 0
40. 3
6. (i) V: (–3, 1), S: Ê - 11 , 1ˆ¯˜ , L.R. = 1, A: y =1 2
ËÁ 4
41. 2 2

(ii) V: (–2, 2).; S: Ê - 5 , 2ˆ¯˜ ; L.R. = 3, A: y = 2 42. –1
ËÁ 4
43. (–2, –9)

Ê 23ˆ 44. x + a + b = 0
ËÁ 4 ¯˜
(iii) V: (2, -6), S: 2, — ; L.R. = 1, A: x = 2 45. 1 a2 (t1 - t2 )(t2 - t3 )(t3 - t1 )
2
Ê 1 1ˆ
(iv) V: ËÁ — 2 , 2¯˜ ; S: (–1/2, 3/4) ; L.R. =1, 46. x2 + y2 – a(1 + t2t3 + t3t1 + t1t2)x

A: 2y – 1 = 0. – a(t1 + t2 + t3 – t1t2t3)y
+ a2(t2t3 + t3t1 + t1t2) = 0
7. (3, 6) and (3, –6)

8. x2 – 2xy + y2 + 32x + 32y + 76 = 0 47. Ê - a, a Ê 1 + 1 + 1 + 1 ˆˆ
ÁË ËÁ m1 m2 m3 m1m2m3 ¯˜ ˜¯
9. x = 0
10. y2 = 8x + 24 = 8(x + 3) 48. x + a = 0
11. y = x2 + 3x + 2
49. (i) 4x + 9 = 0

12. (y – 2)2 = 2(x – 1) (ii) y + 2 = 0
13. (x – 3)2 = 12(y – 1)
(iii) x = 0

14. t1t2 = –1 50. x = 2a + a(t12 + t22 + t1t2 )
15. a ËÁÊ t + 12¯˜ˆ 2
and y = – at1t2(t1 + t2)

16. 4a cosec2 q 51. t2 = - t1 - 2
t1
52. ≥ 8
18. (0, at)

21. (0, 4) 53. 2

22. –3 < x < 3 54. x + y = 3

23. (1/2, –2) or (25/2, –10) 55. y = 2x – 24

24. 16 56. k = 9

25. 4 1 + m2 a(a - mc) 57. 9PQ = 80 10
m2 58. y2 = a(x – 3a)
59. al3 + 2alm2 + nm2 = 0
26. (at1t2, a(t1 + t2))
60. q = tan-1( 2)
27. 2x + y – 4 = 0 and 2x – y + 1 = 0 61. (m2, – 2m) = (1, 2)

28. Ê 1, 2¯˜ˆ or Ê 1, - 2ˆ˜¯ 62. PQ = 6 3
ÁË 2 ËÁ 2 64. y2 = a(x – a)

29. q = p 65. 5
2

30. y = x – 2 and y = –3x + 2 66. 2 5 - 1

31. x + y + a = 0 Ê 2h - 4a 0ˆ¯˜
ÁË 3
y = Ê - a1/3 ˆ x + a Ê - b1/3 ˆ 69. ,
ËÁ b1/3 ¯˜ ËÁ a1/3 ¯˜
32.

4.22 Coordinate Geometry Booster

70. h > 2a Ê p, p˜ˆ¯ Ê p p˜ˆ¯
ËÁ 2 ÁË 2
71. 27ak2 < 4(h – 2a)3 15. , , -

72. j = tan -1 Ê tan q ˆ 17. c > 1/2
ËÁ 2 ¯˜
74. (3a, 0) 18. x – 2y + 8 = 0, (8, 8)

79. 2x – 3y + 4 = 0 19. 12a

80. y = 3x – 3 21. x2 = 2(y – 6)
81. (y2 – 4ax) = (x + a)2 tan2 q 22. (ax + by)(x2 + y2) + (bx – ay)2 = 0

82. 1 ¥ (k 2 - 4ah)(k 2 + 4a2 ) 23. 9y2 = 4a(x – 8a)
|a|
24. y2 = 2a2(x – 4)

83. (k 2 - 4ah)3/2 25. Ê 2 , 8ˆ
2a ËÁ 9 9˜¯

84. 4x – 3y + 1 = 0 15a2
85. y2 = 2a (x – a) 26. 4
86. y2 = 2ax
87. y4 – 2a(x – 2a)y2 + 8a4 = 0 27. a = 2
88. y4 = 2a(x – 4a) 28. (y – 1)2(x + 1) + 4 = 0
89. (2a – b)y2 = 4a2x
90. y2 – 2ax – by + 6ab = 0 30. (3, –4), (0, 0), (8, 16)
91. y2(2x + a) = a(3x + a)2
31. y = - (x - h) - a(y - k)
93. y = a(t1 + t2)
94. 15 ( y - k) (x - h)

95. x – 7y + 10 = 0 32. y2 = a(x – 3a)

LEVEL IV

LEVEL II 1. 8 2 s.u.

1. (c) 2. (c) 3. (b) 4. (c) 5. (a) 2. (c)
6. (d) 7. (c) 8. (c) 9. (a) 10 () 3. (b2 – 4a2)y – 4abx + 4a2b = 0
11. (d) 12. (b) 13. (a) 14. (d) 15. (b)
16. (a) 17. (a) 18. (a) 19. () 20. (d) 4. t=± 1 , j = cos-1 Ê - 1ˆ
21. (c) 22. (b) 23. (b) 24. (b) 25. (b) 5 ÁË 5 ¯˜
26. (b) 27. (b) 28. (d) 29. (d) 30. (a)
31. (b) 32. (b) 33. (d) 34. (c) 35. (a) 5. (a, c)
36. (c) 37. (c) 38. (b) 39. (c) 40. (d)
41. (c) 42. (d) 43. (a) 44. (b) 45. (a) 6. x2 = 2(y – 6)
46. (c) 47. (d) 48. (c) 49. (a) 50. (c) 7. y + 18ËÁÊ x2-16˜¯ˆ 3/2 = 0

LEVEL III 8. x – 2y + 1 = 0
9. (2 – x)y2 = (5y2 + 32)
1. (4, 0) 10. 9y2 = 4ax
2. 4ya2 + 8a3 – xa4 + 4x = 0 11. 9x2 = 4by
3. y2 = 4(a1 – a)(x – a)
4. (0, 0), (4a, 4a), (4a, – 4a), (8a, 0) 12. ( 21 - 5)
5. (x – y)2 = 8(x + y – 2)
8. m = 1, –1 13. 5a
9. (–1, 0) 2
10. (2, 4)
14. |b| < 1
12. y 3 = x + 3 22
13. y = x + 2
14. y = 0, y = 4(x – 1) 15. 4

INTEGER TYPE QUESTIONS

1. 6
2. 3
3. 5

Parabola 4.23

4. 2 7. Ê 2 , 8ˆ
5. 6 ÁË 9 9˜¯
6. 2
7. 6 8. 2 : 1
8. 5
9. 5 9. 15a2
10. 8 4
11. 9
12. 3 10. (b)

COMPREHENSIVE LINK PASSAGES 11. (c)

Passage I: (i) (d) (ii) (c) (iii) (d) 12. (d)
Passage II: (i) (b) (ii) (a) (iii) (c)
Passage III: (i) (d) (ii) (b) (iii) (c) 13. (c)
Passage IV: (i) (a) (ii) (b) (iii) (c)
Passage V: (i) (b) (ii) (c) (iii) (a) 14. (c)
Passage VI: (i) (c) (ii) (b) (iii) (d)
Passage VII: (i) (a) (ii) (c) (iii) (c) (iv) (b) (v) (c) 15. (d)
Passage VII: (i) (c) (ii) (b) (iii) (d)
16. (a)
MATRIX MATCH
17.
1. (A) Æ (S); (B) Æ (R); (C) Æ (Q); (D) Æ (P)
2. (A) Æ (P, R); (B) Æ (P,R); (C) Æ (Q); (D) Æ (Q, S) 18. a = 2
3. (A) Æ (S); (B) Æ (R); (C) Æ (Q); (D) Æ (P)
4. (A) Æ (Q); (B) Æ (P); (C) Æ (Q); (D) Æ (R) 19. (c)
5. (A) Æ (P); (B) Æ (Q); (C) Æ (S); (D) Æ (R) 20. (y – 1)2(x + 1) + 4 = 0
6. (A) Æ (R); (B) Æ (R); (C) Æ (R); (D) Æ (R)
7. (A) Æ (Q); (B) Æ (P); (C) Æ (S); (D) Æ (R) 21. (d)
8. (A) Æ (Q); (B) Æ (S);(C) Æ (P); (D) Æ (R)
22. (a, b)
QUESTIONS ASKED IN PREVIOUS YEARS’ JEE-ADVANCED
EXAMINATIONS 23. (i) Æ A; (ii) Æ B; (iii) Æ D; (iv) Æ C

2. ± 2 24. (i) - (c) (ii)-(b) (iii)-(d)
3. x + y = 3
4. 3/4 25. (a)
5. y2 = 2x – 8
6. (a) 26. (b, c)

27. (a, d)

28. (c, d)

29. (a, b, d)

30. 2

31. 4

32. (b)

33. (d)

34. m = 1

35. (d)

36. (d)

37. (b)

38. 4

39. 2

40. (a, c)

41. (a, b, c)

42. (a, c, d)

HINTS AND SOLUTIONS

LEVEL I a2x2 – 2abxy + b2y2 – 2ax – 2by + 1 = 0
Here A = a2, B = b2 and H = –ab
1. The given equation is Now, H2 – AB = a2b2 – a2b2 = 0
ax + by = 1
Hence, it represents a parabola.
fi ( ax + by )2 = 1 2. Here, a = 1, h = –2 and b = l.

fi ax + by + 2 abxy = 1 Since, it represents a parabola, so
fi (ax + by — 1)2 = (2 abxy )2 h2 – ab = 0
fi a2x2 + b2y2 + 1 + 2abxy – 2ax – 2by
fi 4–l=0
= 4abxy fi l=4

4.24 Coordinate Geometry Booster

3. The given conic is Latus rectum: 4a = 1

16(x2 + ( y - 1)2 ) = (x + 3y - 5)2 Directrix: X + a = 0

fi (x2 + (y - 1)2 ) = 1Ê x + 3y - 5ˆ 2 …(i) fi x+3= 1
4 ËÁ 1+ 3 ¯˜ 4

which represents an ellipse. fi x = - 11
Now, Eq. (i) can also be written as 4

SP2 = e2 ¥ PM2 fi 4x + 11 = 0
Thus, the eccentricity is 1 .
Axis: Y = 0
2 fi y–1=0
fi y=1

4. Let S be the focus, PM be the directrix and the eccen- (ii) The given equation is

tricity = e y2 = 3x + 4y + 2
fi y2 – 4y = x2 + 2
From the definition of conic section, we get fi y2 – 4y + 4 = 2x + 6
fi (y – 2)2 = 3(x + 2)
SP = e fi Y2 = 3X,
PM

fi SP = e ¥ PM where X = x + 2 and Y = y – 2
fi SP2 = e2 ¥ PM2
Vertex: (0, 0)
3y + 10ˆ 2
fi (x - 1)2 + (y - 2)2 = 1 ¥ Ê x + 1 + 9 ¯˜ fi X = 0, Y = 0
2 ËÁ fi x + 2 = 0 and y – 2 = 0

fi 20{(x – 1)2 + (y – 2)2} = (x + 3y + 10)2 fi x = –2 and y = 2
fi 20{x2 + y2 – 2x – 4y +5}
Hence, the vertex is (–2, 2)
= (x2 + 9y2 + 100 + 6xy + 20x + 6y)
Focus: (a, 0)
5. Let S be the focus and PM be the directrix. fi X = a, Y = 0

From the definition of conic section, it is clear that, 3
fi x+2= 4,y–2=0
SP = PM -5
fi SP2 = PM2 fi x= 4 and y = 2

Ê x - y+ 3ˆ 2
ËÁ 1+1 ¯˜
fi ( x - 1)2 + ( y - 1)2 = Hence, the focus is Ê - 5 , 2˜¯ˆ
ÁË 4

fi 2{(x – 1)2 + (y – 1)2} = (x – y + 3)2 Latus rectum: 4a = 3
fi 2(x2 + y2 – 2x – 2y + 2)
Directrix: X + a = 0
= (x2 + y2 + 9 – 2xy –6x – 6y) fi x + 2 = 3/4
fi x2 + 2xy + y2 + 2x + 2y + 4 = 0
fi x = –5/3
6. (i) The given equation is fi 3x + 5 = 0

y2 = x + 2y + 2 Axis: Y = 0
fi y2 – 2y = x + 2
fi y–2=0
fi (y – 1)2 = x + 3 fi y=2
fi Y2 = X, where X = x + 3, Y = y – 1
(iii) The given equation is
Vertex: V(0, 0)
x2 = y + 4x + 2
fi X = 0, Y = 0 fi x2 – 4x = y + 2
fi x2 – 4x + 4 = y + 6
fi x + 3 = 0, y – 1 = 0 fi (x – 2)2 = y + 6
fi X2 = Y,
fi x = –3, y = 1

Hence, the vertex is (–3, 1)

Focus: (a, 0) where X = x – 2 and Y = y + 6

\ X = a, Y = 0 Vertex: (0, 0)
fi X = 0, Y = 0
fi x +3= 1, y -1= 0
4 fi x – 2 = 0 and y + 6 = 0
fi x = 2 and y = –6
fi x = 1 - 3, y = 1
4 Hence, the vertex is (2, –6)

fi x = - 11, y = 1 Focus: (0, a)
4 fi X = 0, Y = a
fi x – 2 = 0 and y + 6 = 1/4
Hence, the focus is Ê - 11 , 1ˆ¯˜ fi x = 2 and y = –23/4
ÁË 4

Parabola 4.25

Hence, the focus is ËÊÁ 2, - 243˜¯ˆ When x = 3, y2 = 12 ¥ 3 = 36
fi y = ±6
Latus rectum: 4a = 1 Hence, the co-ordinates of the points are (3, 6) and
(3, –6).
Directrix: Y + a = 0 8. Let the vertex be V and the focus be S.
The equation of axis is x – y = 0.
fi y + 6 = 1/4
Y
fi 4y + 23 = 0

Axis: X = 0

fi x–2=0

fi x=2

(iv) The given equation is X¢ O
V(–2, –2)
x2 + x + y = 0 X
F(–6, 6)
fi x2 + x = –y
Y¢
ÊÁË 1 ˆ¯˜ 2 1 - ÊËÁ 14¯ˆ˜
2 4
fi x + = - y + = y -

fi X2 = –Y, where

X = x+ 1,Y = y - 1 Let the point Q is the point of intersection of the axis
22
and the directrix.
Vertex: (0, 0) fi X = 0, Y = 0
Clearly, V is the mid-point of Q and S.
fi x + 1 = 0, y - 1 = 0
22 Then Q is (2. 2).

As we know that the directrix is perpendicular to the

fi x= —1, y= 1 axis of the parabola. So, the equation of the directrix is
22
x+y–k=0

Hence, the vertex is Ê - 1 , 1ˆ . which is passing through (2, 2).
Focus : (0, –a) ËÁ 2 2¯˜
Therefore, k = 4.
fi X = 0, Y = –a Hence, the equation of the directrix is

fi x + 1/2 = 0, y – 1/2 = 1/4 x+y–4=0

fi x = –1/2, y = 3/4 Thus the equation of the parabola is

Hence, the focus is (–1/2, 3/4) (x + 6)2 + (y + 6)2 = Ê x + y - 4ˆ
ÁË 1 + 1 ˜¯
Latus rectum: 4a = 1
fi 2((x + 6)2 + (y + 6)2) = (x + y – 4)2
Directrix: Y – a = 0 fi 2(x2 + y2 + 12x + 12y + 36)

fi y – 1/2 – 1/4 = 0 = (x2 + y2 + 16 + 2xy + 8x – 8y)
fi x2 – 2xy + y2 + 32x + 32y + 76 = 0
fi y – 3/4 = 0 9. The given equations are x = t2 + 1 and y = 2t + 1

fi 4y – 3 = 0

Axis: Y = 0 fi y – 1/2 = 0 fi 2y – 1 = 0

7. Let the point be (x, y) Eliminating t, we get
(y – 1)2 = 4(x – 1)
Y
fi Y2 = 4X, where X = (x – 1)

P and Y = (y – 1)

Hence, the equation of the directrix is

OS X X+a=0
fi x–1+1=0
fi x = 0.
10. Let the vertex be V and the focus be S.

Let Q be the point of intersection of the axis and the

directrix.

The given equation is Clearly, Q be (–5, 0) and V be the mid-point of S and Q.
y2 = 12x Then focus S is (–1, 0).

fi 4a = 12 Hence, the equation of the parabola is
fi a = 12/4 = 3
(x + 1)2 + y2 = Ê x + 5ˆ
Given focal distance = 6 ÁË 12 ˜¯
\ x+a=6
fi x+3=6 fi (x + 1)2 + y2 = (x + 5)2
fi x=6–3=3 fi y2 = 8x + 24 = 8(x + 3)

4.26 Coordinate Geometry Booster

11. Let the equation of the parabola be fi t1t2 + 1 = 0

y = ax2 + bx + c …(i) fi t1t2 = –1

which is passing through (0, 2), (–1. 0) and (1, 6). So which is the required relation.

c = 2, a + c = b, a + b + c = 6 15. Since one extremity of the focal chord is P(at2, 2at),

Solving, we get then the other extremity will be Q Ê a , - 2aˆ .
ËÁ t2 t ¯˜
a = 1, b = 3 and c = 2
Hence, the equation of the parabola is y = x2 + 3x + 2. Y
12. Let the equation of the parabola be (y – k)2 = 4a(x – h),

where vertex is (h, k). P(at2, 2at)

Then the equation becomes
(y – 2)2 = 4a(x – 1)

which is passing through (3, 4). X

Therefore, 8a = 4 fi a = 1 O S(a, 0)
2 Q(at12, 2at1)

Hence, the equation of the parabola is Thus, PQ = SP + SQ
(y – 2)2 = 2(x – 1)
= (at 2 + a) + Êa + aˆ˜¯
13. Let the equation of the parabola be (x – H)2 = 4a(y – k), ÁË t2

where vertex is (h, k). = a Ê t 2 + 1 + 2˜¯ˆ = a Ê t + 1ˆ 2
ÁË t2 ÁË t ¯˜
Thus the equation becomes
(x – 3)2 = 4a(y – 2) 16. Now, slope of PQ = 2 1 = tan q
-
Also it given that, the length of the latus rectum = 12 t

fi 4a = 12 fi a = 3.

Hence, the equation of the parabola is
(x – 3)2 = 12(y – 1).

14. Let y2 = 4ax be a parabola, if PQ be a focal chord.

Y t
P
fi 2 cot q = t - 1
t

Ê 1ˆ 2
ËÁ t ¯˜
OS X Thus, PQ = a t +
Q
ÈÎÍÁÊË 1ˆ 2 ˘
t ˜¯ 4˙
= a t - +
˚

Consider any two points on the parabola (at12, 2at1) and = a(4 cot2 q + 4)
(at22, 2at2).
Since PQ passes through the focus S(a, 0), so P, S, Q = 4a cosec2 q
are collinear.
17. S = (a, 0), P = (at 2 , 2at) and Q = Ê a , - 2a ˆ
Thus, m(PS) = m(QS) ÁË t2 t ˜¯

Thus, SP = a + at2, SQ = a + a
t2
fi 2at1 - 0 = 0 - 2at2
at12 - a a - at22 Now, Harmonic mean of SP and SQ

fi 2at1 = 2at2 = 2SP ◊ SQ = 1 2 1
at12 - a at22 - a SP + SQ +

SP SQ

2t1 2t2 = 1 + 1 = 1 + 1 = 1+ t2 = 1
t12 - 1 t22 - 1 SP SQ a + at2 a(1 + t2 ) a
fi = a + a
t2
t1 t2 2 2
fi t12 - 1 = t22 - 1 Thus, 1+1 = 1 = 2a

fi t1(t22 – 1) = t2(t12 – 1) SP SQ a

fi t1t2(t2 – t1) + (t2 – t1) = 0 = semi-latus rectum.

Parabola 4.27
18.
21. Since, the point (l, –l) lies inside of the parabola
Y y2 = 4x, then l2 – 4l < 0
P(at2, 2at) fi l(l – 4) < 0
fi 0<l<4
O S(a, 0) X Hence, the range of l is (0, 4).
Q(at12, 2at1)
22. Since the point (l, 2) is an exterior point of both the
parabolas

Y

P(l, 2)

OX

The equation of the focal chord SP:

y - 0 = 2at - 0(x - a) y2 = (x + 1) and y2 = –(x – 1),
at 2 -a

fi y(t2 – 1) = 2tx – 2at So we have
fi 2tx – (t2 – 1)y – 2at = 0
4 – x – 1 > 0 and 4 + x – 1 > 0
Let d be the distance of the focal chord SP from the fi 3 – x > 0 and 3 + x > 0
vertex (0, 0) to the parabola y2 = 4ax. fi x – 3 < 0 and x + 3 > 0
fi x < 3 and x > –3
Then d = (0 - 0 - 2at) fi –3 < x < 3
4t2 + (t2 - 1)2
23. The given line is

2x + 3y + 5 = 0 …(i)

2at 2a and the parabola is y2 = 8x …(ii)
(t2 + 1)
= = Ê 1ˆ Since (i) is a tangent to the parabola y2 = 8x, so
ÁË t ˜¯
t + ÁËÊ - 2 x - 5 ˜ˆ¯ 2
3
= 8 x

Ê 1ˆ 2 4a2 4a3 fi (2x + 5)2 = 72x
ËÁ t ˜¯ d2 d2 fi 4x2 + 20x + 25 = 72x
Also, PQ = a t + = a ¥ = fi 4x2 – 52x + 25 = 0
fi 4x2 – 2x – 50x + 25 = 0
Thus, PQ a 1
d2 fi 2x(2x – 1) – 25(2x – 1) = 0

Hence the length of the focal chord varies inversely as fi (2x – 1)(2x – 25) = 0

the square of its distance from the vertex of the given fi x = 1/2, x = 25/2

parabola. When x = 1/2, then y = Ê -1 - 5ˆ = -2
ËÁ 3 ¯˜
19. Let the circle described on the focal chord SP, where
S = (a, 0) and P = (at2, 2at). Also, when x = 25/2, then y = –10

The equation of the circle is Hence, the points of contact are
(x – at2)(x – a) + (y – 2at)(y – 0) – 0
(1/2, –2) or (25/2, –10)

Solving it with y-axis, x = 0, we have 24. The given parabola is …(i)
y2 – 2aty + a2t2 = 0 y2 = 12x

Clearly, it has equal roots. fi 4a = 12

So the circle touches the y-axis. fi a=3

Also, the point of contact is (0, at). The given line is 3x + 4y + l = 0

20. The equation of the circle described on AB as diameter fi y=-3x- l …(ii)
44
is

(x - at 2 ) Ê x - a ˆ + (y - 2at) Ê y + 2a ˆ = 0 Since, the line (ii) is a tangent to the parabola (i), so
ËÁ t2 ¯˜ ËÁ t ¯˜
c= a
Put x = a, we have m

Ê 1ˆ Ê 1ˆ 2 fi -l = 3 = -4
ÁË t ˜¯ ËÁ t ˜¯ 4
y2 - 2a t - y + a2 t - = 0 Ê - 3ˆ
ÁË 4¯˜
Clearly, it has equal roots.
Hence the circle touches the directrix at x = –a. fi l = 16

Hence, the value of l is 16.

4.28 Coordinate Geometry Booster

25. Let the equation of the parabola be fi y = 2, 3
Thus, the points are (1, 2) and (1, 3).
y2 = 4ax
and the line be y = mx + c. Hence, the equations of tangents can be at (1, 2) and
(1, 3) be
Solving the above equations, we get
2y = (x + 1) + 5 ( y + 2) - 8
(mx + c)2 = 4ax 2

fi m2x2 + (2mc + 4a)x + c2 = 0 and 3y = (x + 1) + 5 ( y + 3) - 8
2
Let the line y = mx + c intersects the parabola in two
fi 2x + y – 4 = 0 and 2x – y + 1 = 0
real and distinct points, say (x1, y1) and (x2, y2).

Thus (x1 – x2)2 = (x1 + x2)2 – 4x1x2 28. Let the equation of the tangent be

= 4(mc - 2a)2 - 4c2 = 16a(a - mc) , y = 2x + c …(i)
m4 m2 m4
If the equation (i) be a tangent to the parabola, then

and y1 – y2 = m(x1 – x2) c = a = 2 = 1.
Thus, the required length m2

= ( y1 - y2 )2 + (x1 - x2 )2 Thus, the equation of the tangent is

= (x1 - x2 )2 m2 + (x1 - x2 )2 y = 2x + 1 …(ii)
The given parabola is y2 = 8x …(iii)

= 1 + m2 (x1 - x2 ) Solving (ii) and (iii), we get

= 4 1 + m2 a(a - mc) (2x + 1)2 = 8x
m2 fi 4x2 + 4x + 1 = 8x
fi (2x – 1)2 = 0

26. Let the parabola be y2 = 4ax and the two points on the fi x=1
2
parabola are P(at12, 2at1) and Q(at22, 2at2 ) , respec-
tively. When x = 1/2, then y = ± 2
Hence, the point of contacts are
P(t1)
1, 1,
Ê 2 2ˆ¯˜ or Ê 2 - 2˜ˆ¯
ËÁ ÁË

R(at1, at2) a(t1, t2) 29. The equation of line from (–1, 2) is

(y – 2) = m(x + 1)

fi mx – y + (m + 2) = 0

Q(t2) fi y = mx + (m + 2) …(i)

The line (i) will be a tangent to the parabola y2 = 4x, if

The equation of the tangent at P(at12, 2at1) and (m + 2) = 1
Q(at22, 2at2 ) are m

t1y = x + at12 fi m2 + 2m – 1 = 0

…(i) which is a quadratic in m.

and t2 y = x + at2 …(ii) Let its roots are m1, m2.
Solving these equations, we get
Thus, m1 + m2 = –2 and m1m2 = –1
x = at1t2, y = a(t1 + t2) Let q be the angle between them.
Hence the co-ordinates of the point of intersection of
the tangents are (at1t2, a(t1 + t2)). Then, tan (q ) = m2 - m1
1 + m1m2

Notes: = (m1 + m2 )2 - 4m1m2
1 + m1m2
1. x-co-ordinate is the geometric mean of the x-
co-ordinates of P and Q. fi tan (q) = = tan p
2
2. y-co-ordinate is the arithmetic mean of the y-
co-ordinates of P and Q. fi q=p
2
27. The given parabola is y2 = 2x + 5y – 8.
when x = 1, y2 = 5y – 6 Hence, the angle between the tangents is q = p .
fi y2 – 5y + 6 = 0 2

fi (y – 2)(y – 3)

Parabola 4.29

30. Let the equation of the tangent be fi mx2 = 5m2x + 4ab
fi mx2 – 4bm2x – 4ab = 0
(y + 1) = m(x – 1)
Since it has equal roots, so
fi y = mx – (m + 1) …(i)
D=0
Equation (i) be a tangent to the parabola 16b2m4 + 16abm = 0
fi 16bm(bm 3 + a) = 0
y = x2 – 3x + 2, fi m3 = - a
then mx – m – 1 = x2 – 3x + 2
fi x2 – (m + 3)x + (m + 3) = 0 b

Since it has equal roots, so

D=0 fi m = - a1/3
fi (m + 3)2 – 4(m + 3) = 0 b1/3

fi (m + 3)(m + 3 – 4) = 0

fi (m + 3)(m – 1) = 0 Hence, the equation of the common tangent be

fi m = 1, –3

Hence, the equation of the tangents are y = Ê - a1/3 ˆ x + a Ê - b1/3 ˆ
ËÁ b1/3 ¯˜ ÁË a1/3 ¯˜
y = x – 2 and y = –3x + 2

31. The given parabolas are 33. Let the equation of the tangent to the parabola
y2 = 16x is
y2 = 4ax and x2 = 4ay
y = mx + 4
Let the equation of the tangent be y = mx + a . m
m
fi m2x – my + 4 = 0
If it is a tangent to the parabola x2 = 4ay, then …(i)

x2 = 4a ËÊÁ mx + a ˆ˜¯ Y
m

fi mx2 + 4am2x + 4a2 M
fi mx2 – 4am2x – 4a2 = 0

Now D = 0 gives, OX
16a2m2 + 16a2m = 0

fi 16a2m (m3 + 1) = 0
fi m(m3 + 1) = 0

fi m = 0 and –1

Since m = 0 will not satisfy the given tangent, so If the Eq. (i) be a tangent to the circle x2 + y2 = 8, the
m = –1 length of the perpendicular from the centre to the tan-
gent is equal to the radius of the circle.
Hence, the equation of the common tangent be
Therefore,
y = –x – a
fi x+y+a=0 0-0+4 =2 2
32. The given parabolas are y2 = 4ax and x2 = 4by. m4 + m2

Y

P X fi 8(m4 + m2) = 16
fi (m4 + m2 – 2) = 0
O fi (m4 + 2m2 – m2 – 2) = 0
Q fi (m2 + 2)(m2 – 1) = 0
fi (m2 – 1) = 0

fi m = ±1

Let the equation of the tangent be Hence, the common tangents are y = ± x ± 4.
y = mx + a
m 34. Any point on the parabola y = x2 is (t, t2).
Now the tangent at (t, t2) is

…(i) xx1 = 1 (y + y1)
2

Since the equation (i) is also a tangent to the parabola fi tx = 1 ( y + t2 )
x2 = 4by, so 2

x2 = 4b ËÁÊ mx + ma ˜¯ˆ fi 2tx – y – t2 = 0

4.30 Coordinate Geometry Booster

If it is a tangent to the parabola, 37. Any tangent to the parabola y2 = 4x is
y = – (x – 2)2, then
2tx – t2 = – (x – 2)2 y = mx + a
m
fi 2tx – t2 = –x2 + 4x – 4
fi x2 + 2(2 – t)x + (t2 – 4) = 0 fi y = mx + 1
m

Since it has equal roots, fi m2x – my + 1 = 0 …(i)

D=0 If it is a tangent to the circle x2 + (y – 3)2 = 9 the length
4(2 – t)2 – 4(t2 – 4) = 0
fi (2 – t)2 – (t2 – 4) = 0 of the perpendicular from the centre to the tangent is

fi t=2 equal to the radius of the circle.

Therefore,

Hence, the equation of the common tangent is 3m2 + 1 = 3
m4 + m2

y = 4x – 4 fi (3m2 + 1)2 = 9(m4 + m2)
fi (9m4 + 6m2 + 1) = 9(m4 + m2)
35. Let the equation of the tangent to the parabola y2 = 8x is fi 3m2 = 1

y = mx + 2 …(i)
m
Ê 1ˆ
If it is a tangent to the curve xy = –1, then fi m = ± ÁË 3 ˜¯

x Ê mx + 2ˆ = -1 Since, the tangent touches the parabola above x-axis,
ËÁ m ˜¯ so it will make an acute angle with x-axis, so that m is
positive.
fi m2x2 + 2x + m = 0
Thus m = 1 .
Since it has equal roots, so, 3

D=0 Hence, the common tangent is x - 3y + 3 = 0 .
fi 4 – 4m3 = 0
fi m3 = 1 38. The equation of the given parabola is y2 = 4x.
We have, 4a = 4 fi a = 1
fi m=1 Let the end-points of the latus rectum are L(a, 2a) and
L¢(a, –2a).
Hence, the equation of the common tangent is y = x + 2. Therefore L = (1, 2) and L¢ = (1, –2).
As we know that the point of intersection to the tangents
36. Any point on the parabola y2 = x can be considered as at (at12, 2at1) and (at22, 2at2 ) to the parabola y2 = 4ax is
(t2, t).
The equation of the tangent to the parabola y2 = x at
(t2, t) is

yy1 = 1 (x + x1). Ê at1t2 , a Ê t1 + t2 ˆ ˆ
2 ÁË ÁË 2 ˜¯ ¯˜

fi yt = 1 (x + t2 ) Thus, the point of intersection of the tangents at L(1, 2)
2
and L¢(1, –2) is (1, 0).

fi x + 2yt – t2 = 0 39. The equation of the tangent to the parabola

…(i) y2 = 4x …(i)

If it is a tangent to the circle x2 + y2 – 6y + 4 = 0, then from (1, 4) is

(2yt – t2)2 + y2 – 6y + 4 = 0 y – 4 = m(x – 1)
fi y = mx + (4 – m)
fi 4y2t2 + t4 – 4yt3 + y2 – 6y + 4 = 0 …(ii)

fi (4t2 + 1)y2 – 2(2t3 + 3)y + (t4 + 4) = 0 Since (ii) is a tangent to the parabola y2 = 4x, so
c= a
Since it has equal roots, so m

D=0 fi (4 - m) = 1
fi 4(2t3 + 1)2 – 4(4t2 + 1)(t4 + 4) = 0 m
fi (2t3 + 1)2 – (4t2 + 1)(t4 + 4) = 0
fi t4 – 12t3 + 16t2 – 5 = 0 fi 4m – m2 – 1 = 0
fi m2 – 4m + 1 = 0
fi t=1
It has two roots, say m1 and m2.
Hence, the equation of the common tangent is Therefore, m1 + m2 = 4 and m1m2 = 1

x + 2y = 1.

Parabola 4.31

Let q be the angle between the tangents When y = 0, then x = 0.
Then tan (q ) = m2 - m1 Thus, the point on the parabola is (0, 0).
Hence, the required shortest distance
1 + m1m2
= (m1 + m2 )2 - 4m1m2 = 0+0-4 = 4 =2 2
1+1 2
1 + m1m2
42. The given tangents are y + b = m1(x + a)
= 12 = 3 = tan p and y + b = m2(x + a)
23 Therefore, both the tangents pass through (–a, –b)

which is a point lying on the directrix of the parabola.

fi q=p Thus, the angle between them is 90°.

3 p . Hence, the value of m1m2 is –1.
Hence, the angle between the tangents is 43. The equation of the tangent to the curve y = x2 + 6 at
3
(1, 7) is

40. The shortest distance between a line and the parabola 1(y
2
means the shortest distance between a line and a tan- + y1) = xx1 + 6

gent to the parabola parallel to the given line. fi 1 ( y + 7) = x + 6
2
Y

fi 2x – y – 5 = 0 …(i)

Y

X¢ PO X

Q P(1, 7)

Y¢ X¢ O X
Q
Thus, the slopes of the tangent and the line will be the
same. Y¢ …(ii)
Therefore,
The given circle is
2x + 3 = 1
fi x = –1 (x + 8)2 + ( y + 6)2 = ( 100 - c )2
When x = –1, then y = 0.
Hence, the point on the parabola is (–1, 0). If the line (i) be a tangent to the circle (ii), the length of
Thus, the required shortest distance the perpendicular from the centre of the circle is equal
to the radius of the circle.
= -1 - 0 - 2 = 3
1+1 2 Therefore, -16 + 6 - 5 = 100 - c
22 + 1
41. The shortest distance between a line and the parabola
means the shortest distance between a line and a tan- fi 45 = 100 – c
gent to the parabola parallel to the given line.
fi c = 100 – 45 = 5
Y

Thus, the equation of the circle is

Q (x + 8)2 + (y + 6)2 = 45 …(iii)
O fi x2 + y2 + 16x + 12y + 55 = 0

P X Solving Eqs (i) and (iii), we get
x2 + (2x – 5)2 + 16x + 12(2x – 5) + 55 = 0

fi x2 + 4x2 + 25 – 20x + 16x

Thus, the slopes of the tangent and the line will be the + 24x – 60 + 55 = 0
same. fi 5x2 + 20x + 20 = 0
Therefore, - 4 = -1 fi y = 0 fi x2 + 4x + 4 = 0
fi (x + 2)2 = 0
2y + 4
fi x = –2

When x = –2, then y = 2x – 5 = –4 – 5 = –9.

Hence, the point Q is (–2, –9).

4.32 Coordinate Geometry Booster

44. Any tangent to y2 = 4(x + a) is Since it passes through the above three points, we have

y = m1 ( x + a) + a …(i) a2t22t32 + a2 (t2 + t3)2 + 2gat2t3 + 2fa(t2 + t3) + c = 0
m1 …(ii)

Also, any tangent to y2 = 4b(x + b) is a2t12t32 + a2 (t1 + t3)2 + 2gat1t3 + 2fa(t1 + t3) + c = 0
…(iii)
y = m2 (x + b) + b …(ii)
m2 and
a2t12t22 + a2 (t1 + t2 )2 + 2gat1t2 + 2fa(t1 + t2) + c = 0
Since, two tangents are perpendicular, so
…(iv)
m1m2 = –1
Subtracting Eq. (iii) from Eq. (ii) and dividing by a(t2
fi m2 = - 1 – t1), we get
m1
a(t32 (t1 + t2 ) + t1 + t2 + 2t3) + 2gt3 + 2f = 0
From Eq. (ii), we get
similarly from Eqs (iii) and (iv), we get
1
y = - m1 (x + b) - bm1 …(iii) a(t12 (t3 + t2 ) + t3 + t2 + 2t1) + 2gt1 + 2f = 0

Now subtracting Eq. (i) and Eq. (iii), we get From these two equations, we get

m1 ( x + a) + 1 (x + b) + a + bm1 = 0 2g = – a(1 + t2t3 + t3t1 + t1t2),
m1 m1
2f = – a(t1 + t2 + t3 – t1t2t3)

fi Ê m1 + 1ˆ x + Ê m1 + 1ˆ a + Ê m1 + 1ˆ b = 0 Substituting these values of g and f in Eq. (ii), we get
ÁË m1 ¯˜ ËÁ m1 ˜¯ ÁË m1 ¯˜
c = a2(t2t3 + t3t1 + t1t2).

fi x+a+b=0 Thus, the equation of the circle is

Hence, the result. x2 + y2 – a(1 + t2t3 + t3t1 + t1t2)x

45. Let the three points of the parabola be – a(t1 + t2 + t3 – t1t2t3)y

P(at12, 2at1), Q(at22, 2at2 ) and R(at32, 2at3) + a2 (t2t3 + t3t1 + t1t2) = 0

and the points of intersection of the tangents at these 47. Let the equations of the three tangents be

points are A(t2 t3, a(t2 + t3)), B(t1t3, a(t1 + t3)) and y = m1x + a …(i)
A(t1t2, a(t1 + t2) m1 …(ii)
Now, …(iii)

1 at12 2at1 1 y = m2 x + a
2 m2
ar(DPQR) = at22 2at2 1 a
m3
at32 2at3 1 and y = m3 x +

= a2(t1 – t2)(t2 – t3)(t3 – 1) The point of intersection of (ii) and (iii) is

Also, Ê a Ê 1 1 ˆˆ
ËÁ m2m3 ÁË m2 m3 ¯˜ ˜¯
at2t3 a(t2 + t3) 1 , a +
at3t1 a(t3 + t1) 1
ar(DABC) = 1 a(t1 + t2 ) 1 The equation of any line perpendicular to (i) and
2
at1t2 Y

= 1 a2 (t1 - t2 )(t2 - t3 )(t3 - t1)
2
M P(x1, y1)
Hence, the result. TO NQ X
46. Let P, Q and R be the points on the parabola y2 = 4ax at

which tangents are drawn and let their co-ordinates be

P(at12, 2at1), Q(at22, 2at2 ) and R(at32, 2at3). The equation of any tangent to the parabola at (x1, y1) is
The tangents at Q and R intersect at the point (y – y1) = m(x – x1),

A[at2t3, a(t2 + t3)]

Similarly, the other pairs of tangents at the points where m = Ê dy ˆ
ËÁ dx ˜¯
B[at1t3, a(t1 + t3)] and C[at1t2, a(t1 + t2)] ( x1, y1)

Let the equation to the circle be passes through the point of intersection of tangents

x2 + y2 + 2gx + 2fy + c = 0 …(i) (ii) and (iii) is

Parabola 4.33

y - a Ê 1 + 1ˆ = -1 Ê x - aˆ We have,
ËÁ m2 m3 ¯˜ m1 ËÁ m2m3 ˜¯ 4a = 1

i.e. y + x = a È 1 + 1 + 1˘ …(iv) fi a=1
m1 ÍÎ m2 m3 m1m2m3 ˙˚ 4

Hence, the equation of the director circle is

Similarly the equation to the straight line through the X+a=0

point of intersection of (iii) and (i) and perpendicular fi x+2+ 1 =0
4
to (ii) is

y + x = a È 1 + 1 + 1˘ …(v) fi 4x + 9 = 0
m2 ÎÍ m3 m1 m1m2m3 ˚˙
(ii) The given parabola is
and the equation of the straight line through the point of
x2 = 4x + 4y
intersection of (i) and (ii) and perpendicular to (iii) is fi (x2 – 4x + 4) = 4y + 4 = 4(y + 1)
fi (x – 2)2 = 4(y + 1)
y + x = a È 1 + 1 + 1˘ …(vi) fi X2 = 4Y,
m3 ÎÍ m1 m2 m1m2m3 ˙˚
where X = x – 2

The point which is common to the straight lines (iv), and Y = y + 1

(v) and (vi), i.e. the orthocentre of the triangle is We have, 4a = 4

Ê Ê 1 1 1 1 ˆˆ fi a=1
ËÁ ËÁ m1 m2 m3 m1m2m3 ¯˜ ¯˜
- a, a + + + Hence, the equation of the director circle is

Y+a=0

Hence, the point lies on the directrix. fi y+1+1=0

48. fi y+2=0

Y (iii) The given parabola is

P(at12, 2at1) y2 = 4x + 4y – 8
X fi y2 – 4y + 4 = 4x – 8 + 4
fi (y – 2)2 = 4x – 4 = 4(x – 1)
O fi Y 2 = 4X,

where X = (x – 1)

and Y = (y – 2)

We have 4a = 4 fi a = 1

x+a=0 Hence, the equation of the director circle is

Q(at22, 2at2) X+a=0

fi x–1+1=0

The equations of tangents at P and Q are yt1 = x + at12 fi x=0
and yt2 = x + at2. 50. Let the parabola be y2 = 4ax

The point of intersection of these tangents is and two points on the pa- P

[at1t2, a(t1 + t2)] rabola are P(at12, 2at1) and A
Let this point be (h, k). Q(at22, 2at2 ) .
The equation of the nor-
1 1
The slope of the tangents are m1 = t1 and m2 = t2 . mals at P(at12, 2at1) and Q
Q(at22, 2at2 ) are
Since these two tangents are perpendicular, so
m1m2 = –a y = - t1x + 2at1 + at13 …(i)
and

fi 1 ◊ 1 = -1 y = - t2x + 2at2 + at23 …(ii)
t1 t2
Solving Eqs (i) and (ii), we get
fi t1 ◊ t2 = –1
fi h = –a x = 2a + a(t12 + t22 + t1t2 ) P
Q
Thus the locus of the points of intersection is x + a = 0 and y = –at2t2(t1 + t2).
which is the directrix of the parabola y2 = 4ax. 51. Let the parabola be y2 = 4ax

49. (i) The given parabola is and the two points on the pa-
y2 = x + 2 rabola are P(at12, 2at1) and
Q(at22, 2at2 ) .
fi Y 2 = X,

where X = x + 2 and Y = y

4.34 Coordinate Geometry Booster

The equation of the normal to the parabola at The line (i) will be a normal to the given parabola, if
k = –2am – am3 = 6 + 3 = 9.
P(at12, 2at1) is y = - t1x + 2at1 + at13 which meets the Hence, the value of k is 9.
parabola again at Q(at22, 2at2 ) . 57. The equation of the parabola is y2 = 8x.
Thus, 2at2 = - at1t22 + 2at1 + at13 We have, 4a = 8 fi a = 2

fi 2a(t2 - t1) + at1(t22 - t12 ) = 0 Y

fi (t2 – t1)[2a + at1(t2 + t1)] = 0 P

fi [2 + t1(t2 + t1)] = 0

fi t12 + t1t2 + 2 = 0 X
O
fi t2 = - t1 - 2
t1

which is the required condition.

52. As we know that if the normal at t1 meets the parabola Q
again at t2, then
The equation of the normal to the given parabola at

t2 = - t1 - 2 P(8, 12) is
t1
y - y1 = - y1 (x - x1)
2 2a
Ê 2 ˆ
fi t22 = ÁË - t1 - t1 ¯˜ fi y - 12 = - 12 (x - 18)
4
= t12 + 4 + 4≥ 4+ 4=8
t12 fi y – 12 = –3x + 54
fi 3x + y = 66
Hence, the result. Q Solving y = 66 – 3x and the parabola y2 = 8x, we get

53. Since the normal at t1 P x = - 44 and y = 242
meets the parabola at 39

t3, so t3 = - t1 - 2. Ê 44 242ˆ
t1 ËÁ 3 9 ˜¯
Hence, the co-ordinates of Q is - , .

Similarly, t3 = -t2 - 2 2 2
t2
R Thus, PQ = Ê 242 18˜ˆ¯ Ê 44 12¯˜ˆ
ËÁ 9 ÁË 3
2 2 - + - -
t1 t2
Thus, - t1 - = -t2 -

fi (t1 - t2 ) = Ê 2 - 2 ˆ = 2(t1 - t2 ) = 6400 + 6400
ÁË t1 t2 ¯˜ t1t2 81 9

fi t1t2 = 2 = 80 1 + 1
81 9
54. The equation of the normal is
= 80 ¥ 10
y - y1 = - y1 (x - x1) , where a = 1 9
2a
fi 9PQ = 80 10

fi y - 2 = - 2 (x - 1) = - x + 1 58. The equations of normals at P(t1) and Q(t2) are
2 y = –t1x + 2at1 + at13

fi x+y=3 and y = –t2x + 2at2 + at23

55. The equation of the normal to the parabola y2 = 4ax is P
y = mx – 2am – am3

Here, a = 2 and m = 2.

Therefore, y = 2x – 8 – 16 = 2x – 24 RX
O
Hence, the equation of the normal is y = 2 x – 24.
56. The given parabola is y2 = 12x. Q

We have, 4a = 12 fi a = 3. Since these two normals are at right angles, so t1t2 = –1.

The given line x + y = k fi y = –x + k …(i)

Parabola 4.35

Let M(h, k) be the point of intersection of two normals. 62. If the normal at P(t1) meets the parabola at Q(t2), then
Then, h = 2a + a(t12 + t1t2 + t22 )
t2 = - t1 - 2 …(i)
and k = – at1t2(t1 + t2) t1
fi h = 2a + a {(t1 + t2)2 – 2t1t2}
Y
and k = –at1t2(t1 + t2)
fi h = 2a + a{(t1 + t2)2 + 2} P

and k = a(t1 + t2) OX

Eliminating t1 and t2, we get,
k2 = a(h – 3a)

Hence, the locus of M(h, k) is y2 = a(x – 3a).

59. The given line is

lx + my = 0 Q
fi my = –lx – n
Since the normal chord subtends an angle of 90° at the
fi y = Ê - lˆ x + Ê - nˆ vertex, then
ÁË m ˜¯ ËÁ m ¯˜
t1t2 = –4
As we know that the line y = mx + c will be a normal to From Eq. (i), we get
the parabola y2 = 4ax if
t12 + t1t2 + 2 = 0
c = –2am – am3
fi t12 – 4 + 2 = 0
ËÊÁ - mn ˆ˜¯ Ê lˆ Ê l ˆ3
fi = - 2a ÁË - m ˜¯ - a ËÁ - m ¯˜ fi t12 – 2 = 0

Ê 2 ˆ 2
ËÁ t1 ¯˜
fi al3 + 2alm2 + nm2 = 0 Also, t22 = -t1 -

Hence, the result. = t12 + 4 + 4
60. The equation of the parabola is y2 = 4ax. t12

If the normal at P(t1) meets the parabola again at Q(t2), =2+2+4=8
then

t2 = - t1 - 2 Therefore,
t1
PQ2 = a2(t12 – t22)2 + 4a2(t1 – t2)2

fi t1t2 = –t12 – 2 = 1◊ (2 - 8)2 + 4( 2 + 2 2)2

fi t12 + t1t2 + 2 = 0 …(i) = 36 + 72 = 108

The chord joining t1, t2 subtends a right angle at the fi PQ = 6 3
vertex, so the product of their slopes = –1 63. The equation of the parabola is y2 = 4ax.

fi 2 ◊ 2 = -1 Y
t1 t2

fi t1t2 = –4 …(ii) P

From Eqs (i) and (ii), we get

t12 – 4 + 2 = 0 X
fi t12 = 2 O

fi t1 = 2 Q

fi tan q = 2 Let the normal chord be PQ, where P(t1) and Q(t2).
fi q = tan-1( 2) Since the abscissa and ordinate of the point (p, p) are
61. The given equation of the parabola is y2 = 4x.
same, then
We have 4a = 4 fi a = 1.

The equation of the normal to the parabola 2at1 = at12

y2 = 4x at (am2, –2am) is fi t1 = 2
y = mx – 2am – am3 = mx – 2m – m3
If the normal at P(t1) meets the parabola Q(t2), then
Since, the normal makes equal angles with the axes, so
t2 = - t1 - 2
m = ±1 t1
Thus, the points are (m2, –2m) = (1, 2)
fi t2 = –2 – 1 = –3

4.36 Coordinate Geometry Booster

Let S(a, 0) be the focus of the parabola y2 = 4ax. fi m3 – 4m = 0
fi m = 0, –2, 2
Then the slope of SP = 2at1 = 4= 4 Therefore, P is (4, –4) or (4, 4) and let C(6, 0) be the
at12 - a 4 -1 3 centre of the circle and Q be a point on the circle.
Therefore,
and the slope of SQ = 2at2
at22 - a CP = (6 - 4)2 + (4 - 0)2 = 20 = 2 5

= —6 = —6 = - 3 and CQ = 5
9-1 8 4 Thus, the shortest distance = CP – CQ

It is clear that = 2 5- 5= 5
66. The given equation of the parabola is y2 = 8x.
m(SP) ¥ m(SQ) = 4 ¥ - 3 = -1
34 We have, 4a = 8 fi a = 2.

Hence, the result. Y
64. The given equation of the parabola is y2 = 4ax.
…(i)
The equation of the normal at P(am2, –2am) is

y = mx – 2am – am3

Let Q be a point on the axis of the parabola. O X
P
Put y = 0 in Eq. (i), we get
x = 2a + am2 Q

Hence, the co-ordinates of the point Q is (2a + am2, 0). O

Let M(h, k) be the mid-point of the normal PQ. The equation of the normal to the parabola …(i)
y2 = 8x at P(4m2, –4m) is
Then, h = am2 + 2a + am2 y = mx – 4m – 2m3
2
The given equation of the circle is
and k = - 2am = - am x2 + y2 + 12y + 35 = 0
2
fi x2 + (y + 6)2 = 1
Eliminating m, we get

h = a + k2
a

fi a2 + k2 = ah

Hence, the locus of M(h, k) is Thus, the centre of the circle is C(0, –6).
a2 + y2 = ax

fi y2 = a(x – a) As we know that the shortest distance between a circle
65. The given equation of the parabola is y2 = 4x.
and the parabola lies along the common normal.
The equation of the normal at P(m2, –2m) to the
Therefore, the normal always passes through the centre
parabola y2 = 4x is
of the circle. So
y = mx – 2m – m3 …(i) –6 = –4m – 2m3

The given equation of the circle is fi m3 + 2m – 3 = 0

x2 + y2 – 12x + 31 = 0 …(ii) fi m=1
fi (x – 6)2 + y2 = 5
Thus, the point P is (4, –4).

Y Let Q be any point on the circle.

Then CQ = 1 and

P CP = (4 - 0)2 + (- 4 + 6)2 = 20 = 2 5
Q
X Hence, the shortest distance = PQ
OC
= CP – CQ = 2 5 - 1 .
The shortest distance between the parabola and the 67. The equation of any normal to a parabola y2 = 4ax is
circle lies along the common normal.
y = mx – 2am – am3
Therefore, the centre of a circle passes through the nor-
mal, so we have which meets at a point, say (h, k).
Thus, am3 + (2a – h)m + k = 0.
0 = 6m – 2m – m3
which is a cubic equation in m.

So it has three roots, say m1, m2 and m3.
Therefore, m1 + m2 + m3 = 0
Hence the result.

Parabola 4.37

68. Let the ordinates of A, B and C be y1, y2 and y3 respec- fi k 2 - 4(h - 2a)3 < 0
tively. 27a

Then, y1 = –2am1, y2 = –2am2, y3 = –2am3. fi 27ak2 < 4(h – 2a)3
Thus,

y1 + y2 + y3 = –2am3 – 2am2 – 2am3 Hence, the result.

= –2a(m1 + m2 + m3) = –2a.0 = 0 72. Let the normal at P(at12, 2at1) be y = –t1x + 2at1 + at13

69. If A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of a Thus slope of the normal = tan q = –t1
DABC, then its centroid is
It meets the parabola again at Q(at22, 2at2)
x1 + x2 + x3 , y1 + y2 + y3 ˜¯ˆ
ÁÊË 3 3 Then t2 = - t1 - 2 .
t1
Ê x1 + x2 + x3 0¯ˆ˜
= ËÁ 3 , Now the angle between the normal and the parabola =

angle between the normal and the tangent at Q. If j be

Hence the centroid lies on the axis of the parabola. the angle between them, then

Also, tan j = m1 - m2
1 + m1m2
ÊÁË x1 + x2 + x3 ¯ˆ˜ = 1 (am12 + am22 + am32 )
3 3 1
-t1 - t2
a ÏÌ0 2 ÊËÁ 2a - h ˆ¯˜ ¸ 2h - 4a
= 3 Ó - a ˝ = 3 =
˛
1 + (- t1 ) Ê 1 ˆ
Ê 2h - 4a 0˜¯ˆ ÁË t2 ¯˜
Thus, the centroid of DABC is ËÁ 3 , .
= - t1t2 + 1
70. When three normals are real, then all the three roots of t2 - t1
am3 + (2a – h)m + k = 0 are real.
t1 Ê -t1 - 2 ˆ + 1
Let its three roots are m1, m2, m3. ÁË t1 ˜¯ t1
For any real values of m1, m2, m3, = - -
2
m12 + m22 + m32 > 0 t1 - t1 -
fi (m1 + m2 + m3)2 – 2(m1m2 + m2m3 + m3m1) > 0
= - - t12 - 1
2 ÊÁË 2a - h ˜¯ˆ Ê 1 + t12 ˆ
fi 0 - a > 0 - 2 ËÁ t1 ˜¯

fi h – 2a > 0 = - t1
2
fi h > 2a
71. Let f(m) = am3 + (2a – h)m + k = tan q
2
fi f ¢(m) = 3am2 + (2a – h) 2a

If f(m) has three distinct roots, so f ¢(m) has two distinct fi j = tan -1 Ê tan q ˆ
roots. ËÁ 2 ¯˜
Let two distinct roots of f ¢(m) = 0 are a and b.

Thus, a = Ê h — 2aˆ and b = - ÊËÁ h — 2a ˆ˜¯ . 73. The equation of the normal to the parabola y2 = 4ax at
ÁË 3 ˜¯ 3 P(at2, 2at) is
y = –tx + 2at + at3

Now, f(a)f(b) = 0 Y

f(a) f(–a) = 0

fi (aa3 + (2a – h)a + k)(–aa3 – (2a – h)a + k) < 0 P
O
fi k2 – ((aa2 + (2a – h))2 a2) < 0 S X

fi k2 - Ê h - 2a + (2a - h)˜ˆ¯ 2 Ê h - 2aˆ < 0
ÁË 3 ËÁ 3 ¯˜

Ê 4a - 2h ˆ 2 Ê h - 2aˆ Q
ËÁ 3 ˜¯ ËÁ 3 ¯˜
fi k2 - < 0

4.38 Coordinate Geometry Booster

It is given that at2 = 2at fi t = 2. Thus, SP = PM = x1 + a,
SG = AG – AS = x1 + 2x – a = x1 + a.
Thus, the co-ordinates of P is (4a, 4a) and focus is and ST = AS +AT = a + x1.
Hence, SP = SG = ST.
S(a, 0). 76. The normal at P(t) is y = –tx + 2at + at3.

Also, the normal chord meets the parabola at some Y

point, say Q. Then the co-ordinates of Q is (9a, –6a).

Now,

Slope of SP = m1 = 4a = 4
3a 3
P
-6a = -3.
and the slope of SQ = m2 = 8a 4

4 4 ONG X
3 3
Thus, m1 ¥ m2 = ¥ - = -1

Hence, the result. It meets the x-axis at G.
74. Let the latus rectum be LSL¢, where L = (a, 2a) and Thus the co-ordinates of G be (2a + at2, 0).
Also N is (at2, 0).
L¢ = (a, –2a). Thus NG = 2a + at2 – at2 = 2a = semi-latus rectum.
77. The normal at P(t) is y = –tx + 2at + at3
Y
Y
L
M B

OS P(3a, 0) X

Z L¢

OS G

Normal at L(a, 2a) is x + y = 3a …(i)
Normal at L¢(a, –2a) is x – y = 3a …(ii)
Clearly, (ii) is perpendicular to (i).
Solving, we get Thus, S is (a, 0), G is (2a + at2, 0) and P is (at2, 2at).
Now, SP = a + x = a + at2 = a(12 + t2)
x = 3a and y = 0
Hence, the point of intersection is (3a, 0). SG = 2a + at2 – a = a + at2 = a(1 + t2) = SP
75. Thus, P and G are equidistant from the focus.
78.
Y
Y
P(x1, y1) Q

T O S(a, 0) N GX P

O GX

Let P(x1, y1) be any point on the parabola y2 = 4ax. The normal at P(t) is y = –tx + 2at + at3
The equation of any tangent and any normal at P(x1, y1)
are Thus G is (2a + at2, 0) and P is (at2, 2at).
Now, PG2 = 4a2 + 4a2t2
yy1 = 2a(x + x1) and y - y1 = - x - x1 …(i)
y1 2a
Q is a point on the parabola such that QG is perpen-

Since the tangent and the normal meet its axis at T and dicular to axis so that its ordinate is QG and abscissa is

G, respectively, so the co-ordinates of T and G are (–x1, the same as of G.
0) and (x1 + 2a, 0), respectively. Hence, the point Q is (2a + at2, QG).

Parabola 4.39

But Q lies on the parabola y2 = 4ax. Now,
QR = (at12 - at22 )2 + (2at1 - 2at2 )2
Now,
QG2 = 4a(2a + at2) = |a(t1 - t2 )| (t1 + t2 )2 - 4
= 8a2 + 4a2t2
= (4a2 + 4a2t2) + 4a2 = |a| Ê k2 - 4h ˆ Ê k2 ˆ
= PG2 + 4a2 ËÁ a2 a ¯˜ ËÁ a2 + 4˜¯

fi QG2 – PG2 = 4a2 = constant.

Hence, the result. = 1 ¥ (k 2 - 4ah)(k 2 + 4a2 )
|a|
79. The equation of the chord of contact of the tangents
from the point (2, 3) to the parabola y2 = 4x is 83. Let tangents are drawn from P(h, k) to the parabola
y2 = 4ax, intersects the parabola at Q and R.
yy1 = 2(x + x1)
fi 3y = 2(x + 2)

fi 2x – 3y + 4 = 0 Q

80. The equation of the chord of contact of the tangents to
the parabola y2 = 12x drawn through the point (–1, 2) is

2y = 6(x – 1) P
R
fi y = 3x – 3
81. Let the point of intersection of the tangents be R(a, b). Then the chord of contact of the tangents to the given
parabola is QR.
The equation of the tangent to the parabola at Then QR is
P(at12, 2at1) and Q(at22, 2at2 ) are
Thus, the slopes of the tangents are yk = 2a(x + h)
fi 2ax – yk + 2ah = 0
m1 = 1 and m2 = 1
t1 t2 Therefore PM = the length of perpendicular from
P(h, k) to QR
Then the point of intersection of the two tangents be
[at1t2, a(t1 + t2)]. = 2ah - k 2 + 2ah
Therefore a = at1t2 and b = a(t1 + t2). k 2 + 4a2
Let q be the angle between the two tangents. Then

1-1
tan (q ) = t1 t2 = t2 - t1
1+ 1 t1t2 + 1 = - k 2 - 4ah
k 2 + 4a2
t1t2

fi (1 + t1t2 ) tan (q ) = ( (t1 + t2 )2 - 4t1t2 ) = k 2 - 4ah
k 2 + 4a2

fi (1 + t1t2)2 tan2 q = (t1 + t2)2 – 4t1t2 Thus, the area (DPQR)

ËÁÊ1 + a ˜ˆ¯ 2 b2 4a b 2 - 4aa = 1 ◊ QR ◊ PM
a a2 a a2 2
fi tan 2q = - =

fi (a + a)2 tan2 q = (b2 – 4aa) = 1 ¥ 1 (k 2 - 4ah)(k 2 + 4a2 ) ¥ (k 2 - 4ah)

Hence, the locus of (a, b) is 2 |a| k 2 + 4a2

(y2 – 4ax) = (x + a)2 tan2 q = (k 2 - 4ah)3/2 if a > 0
2a
82. The equation of the parabola is y2 = 4ax and the point y2 = 4x,
84. The equation of the chord of the parabola
(h, k) be P. which is bisected at (2, 3) is

Let the tangents from P touch the parabola at T = S1
fi yy1 – 2a(x + x1) = y12 – 4ax1
Q(at12, 2at1) and R(at22, 2at2 ) , then P is the point of fi 3y – 4(x + 2) = 9 – 8.2 = 7.
intersection of the tangents.
fi 3y – 4x – 1 = 0
Therefore, h = at1t2 and k = a(t1 + t2)
fi 4x – 3y + 1 = 0
fi t1t2 = h and (t1 + t2 ) = k
a a

4.40 Coordinate Geometry Booster

85. Let the equation of the parabola be y2 = 4ax. 88. Let QR be the chord and M(a, b) be the mid-point of it.

The equation of the chord of the parabola, whose Y
mid-point (x1, y1) is A

T = S1. OM X
fi yy1 – 2a(x + x1) = y12 – 4ax1
If it is a focal chord, then it will pass through the focus
(a, 0) of the parabola.

Therefore, B

0. y1 – 2a(x + x1) = y12 – 4ax1 Then the equation of the chord of a parabola
fi y12 = –2a2 – 2ax1 + 4ax1 = 2a(x1 – a) y2 = 4ax at M(a, b) is
Hence, the locus of (x1, y1) is y2 = 2a(x – a).
86. Let the equation of the parabola be y2 = 4ax. T = S1
fi yb – 2a (x + a) = b2 – 2aa
Let VP be any chord of the parabola through the vertex fi yb – 2ax = b2 – 2aa …(i)
and M(h, k) be the mid-pont of it.

Then, the co-ordinates of P becomes (2h, 2k). Let V(0, 0) be the vertex of the parabola.

Since P lies on the parabola, so The combined equation of VQ and VR, making homo-
(2k)2 = 4a ◊ (2h)
geneous by means of (i), we have
fi 4k2 = 8ah
fi k2 = 2ah y2 = 4ax ¥ Ê yb - 2ax ˆ
Hence, the locus of (h, k) is y2 = 2ax. ÁË b2 - 2aa ˜¯

fi y2(b2 – 2aa) – 4abxy + 8a2x2 = 0

87. Equation of the normal at any point (at2, 2at) of the Since, the chord QR subtends a right angle at the ver-
parabola y2 = 4ax is
tex, so we have
y = –tx + 2at + at3 …(i)
co-oefficient of x2 +co-efficient of y2 = 0
fi (b2 – 2aa) + 8a2 = 0

Lep PQ be the normal, whose mid-point is M(a, b). fi b2 = 2a(a – 4a)

Therefore, Hence, the locus of M(a, b) is

T = S1 y2 = 2a(x – 4a)
fi yb – 2a(x + a) = b2 – 3aa
89. Let QR be the chord of the parabola and M(a, b) be its
fi yb = 2a(x + a) = (b2 – 4aa)
…(ii) mid-point. Then the equation of the chord QR bisected
at M(b, b) is
Equations (i) and (ii) are identical.
T = S1
Therefore, 1 = t= 2at + at3 fi yb – 2a(x + a) = b2 – 4aa
b —2a b 2 - 2aa
fi yb = 2a(x + a) + (b2 – 4aa)

t = - 2a t 2at + at3 fi yb = 2ax + (b2 – 2aa) …(i)
b —2a b 2 - 2aa
fi and = If the Eq. (i) be a tangent to the parabola y2 = 4bx, then

From the above two relations, eliminating t, we get c= b
m

Ê - 2a ˆ 2a Ê - 2aˆ + a Ê - 2a ˆ 3 fi Ê b2 - 2aa ˆ = b = bb
ËÁ b ˜¯ ÁË b ˜¯ ËÁ b ˜¯ ÁË b ˜¯ (2a /b ) 2a
=
—2a b 2 - 2aa fi (b – 2a)b2 + 4a2a = 0

fi (b 2 - 2aa ) = - 2a Ê 2a + a Ê - 2a ˆ 2ˆ Hence, the locus of M(a, b) is
Á ËÁ b ˜¯ ˜ (2a – b)y2 = 4a2x
Ë ¯
90. Let the mid-point of the chord be M(h, k).
fi b2(b2 – 2aa) = –2a(2ab2 + 4a3)
fi b2(b2 – 2aa) = –4a2b2 – 8a4) Then the equation of the chord at M(h, k) is
fi b4 – 2a(a – 2a)b2 + 8a4 = 0
T = S1
Hence the locus of M(a, b) is fi yk – 2a(x + h) = k2 – 4ah
y4 – 2a(x – 2a)y2 + 8a4 = 0
which passes through the point (3b, b).
Then, bk – 2a(3b + h) = k2 – 4ah.
fi k2 – 2ah – bk + 6ab = 0

Parabola 4.41

Hence, the locus of M(h, k) is Now,
y2 – 2ax – by + 6ab = 0 Slope of AB

91. The equation of the given parabola is …(i) = m( AB) = 2a(t2 - t1) = t1 2
y2 = 4ax …(ii) a(t22 - t12 ) + t2
…(iii)
The equation of the tangent at P (at2, 2at) is The equation of the diameter is
yt = x + at2
y = 2a fi y = a(t1 + t2 ) …(i)
The equation of the directrix of the parabola m
y2 = 4ax is x + a = 0 Now the tangents at A = (at12, 2at1) and B = (at22, 2at2 )

Solving Eqs (ii) and (iii), we get meet at N[at1t2, a(t1 + t2)].

x = –a and y = a(t2 - 1) Thus, N lies on y = a(t1 + t2).
t
94. As we know that all rays of light parallel to x-axis of

Thus, the point on the directrix, say Q, whose co-ordi- the parabola are reflected through the focus of the pa-

Ê a(t2 - 1)ˆ rabola.
ËÁ - a, t ˜¯
nates are . The equation of the given parabola is
(y – 4)2 = 8(x + 1)
Let M(h, k) be the mid-point of P and Q. Then
fi Y2 = 8X,
h = at2 - a and k = a(t2 - 1) + 2at
2 2t 2 where Y = y – 4 and X = x + 1

fi t2 = 2h + a and 4k2t2 = a2(3t2 – 1)2 Now the focus of the parabola is (a, 0).
a
Therefore,

Eliminating t, we get X = a and Y = 0

Ê 2h + aˆ Ê 3ÁËÊ 2h + aˆ - 1ˆ¯˜ 2 fi x + 1 = 2 and y – 4 = 0
ËÁ a ˜¯ ËÁ a ¯˜
4k 2 = a2 fi x = 1 and y = 4

fi 4k2(2h + a) = a(6h + 3a – a)2 Hence, the focus is (1, 4).
fi k2(2h + a) = a(3h + a)2
Thus a = 1 and b = 4

Hence, the locus of M(h, k) is Now, a + b + 10 = 1 + 4 + 10 = 15.
y2(2x + a) = a(3x + a)2
95. Let the line y = x + 2 intersects the parabola at P.
92. Let y = mx + c represents the system of parallel chords. Solving the line y = x + 2 and the parabola y2 = 4(x + 2),
The equation of the diameter to the parabola y2 = 4ax is we get

y = 2a . the point P is (2, 4).
m
a 2aˆ Now the equation of the tangent to the parabola
The diameter meets the parabola y2 = 4ax at Ê m2 , m ¯˜ . y2 = 4(x + 2) at P(2, 4) is
ÁË
yy1 = 2(x + x1) + 8
The equation of the tangent to the parabola y2 = 4ax at fi 4y = 2(x + 2) + 8

Ê a , 2a ˆ is y = mx + a . fi x – 2y + 6 = 0
ÁË m2 m ¯˜ m
Let IP be the incident ray, PM be the reflected ray and
which is parallel to y = mx + c. A PN be the normal
93. As we know that the normal is equally inclined with the
incident ray as well as the reflected ray.
Now the slope of IP = 1, slope of normal PN = –2 and
let the slope of the reflected ray = m. Then

NO M 1+ 2 = -2 - m
1 - 2 1 - 2m

fi m=1
7

B Hence, the equation of the reflected ray is

Let AB be the chord, where A = (at12, 2at1) and y - 2 = 1 (x - 4)
B = (at22, 2at2 ) . 7

fi 7y – 14 = x – 4

fi x – 7y + 10 = 0

4.42 Coordinate Geometry Booster

LEVEL III Now, m(OA) ¥ m(OC) = –1

fi 2at ¥ - 2at = -1
at 2 at 2
1. Clearly, focus = Ê 3 - 5 , 6+ 6ˆ = (-1, 6)
ËÁ 2 2 ¯˜
fi 2t ¥ -2t = -1
t2 t2
2. Given y = a2x2 + a2x - 2a
32 fi t2 = 4

fi t=2

fi 6y = 2a2x2 + 3a2x – 12a Hence, the co-ordinates of the vertices are O = (0, 0);

A = (4a, 4a); B = (8a, 0), and C = (4a, –4a).

fi 2a2 Ê x2 + 3 x¯˜ˆ = 6y + 12a = 6( y + 2a) 5. Let P = (x, y).
ÁË 2
Y

fi Ê x2 + 3 x¯ˆ˜ = 3 (y + 2a)
ÁË 2 a2
P S(2 2, 2 2)

Ê 3ˆ 2 3 9
ËÁ 4 ¯˜ a2 16
fi x + = (y + 2a) + M V( 2, 2)
y=x x+y=0
X¢ X

Ê 3ˆ 2 3 Ê 3a2 ˆ
ËÁ 4¯˜ a2 ÁË 16 ¯˜
fi x + = y + 2a +

3 Y¢
a2
fi X 2 = Y , We have,

where X = x + 3 , Y = y + 2a + 3a2 SP = PM
4 16 fi SP2 = PM2

Ê x +y ˆ 2
ËÁ 2 ˜¯
Let focus = (h, k) fi (x - 2 2)2 + (y - 2 2)2 =

X = 0 and Y = 3 fi 2[(x - 2 2)2 + ( y - 2 2)2 ] = (x + y)2
4a2
3a2
fi x + 3 = 0 and y + 2a + 16 = 3 fi (x - y)2 = 8 2(x + y - 2 2)
4 4a2

fi h= — 3, k + 2a + 3a2 = 3 6. Clearly, the point of intersection is (4a, 4a) which satis-
4 16 4a2 fies the straight line
2bx + 3cy + 4d = 0
fi k + 2a + — 4ha 2 = — 4h fi 2b(4a) + 3c(4a) + 4d = 0
16 4a2 fi 2ab + 3ac + d = 0
fi a(2b + 3c) + d = 0
fi 16ka2 + 32a3 – 4ha4 + 16h = 0 fi a(2b + 3c) = –d
fi 4ka2 + 8a3 – ha4 + 4h = 0 fi a2(2b + 3c)2 = d2

Hence, the locus of (h, k) is Ê d ˆ 2
4ya2 + 8a3 – xa4 + 4x = 0 ÁË a ¯˜
fi (2b + 3c)2 =
3. Clearly, the equation of the parabola is
y2 = 4(a1 – a)(x – a) fi d2 = a2(2b + 3c)2
7. Any tangent to the parabola y2 = 4a(x + a) is
4. Let A = (at2, 2at), C = (at2, – 2at)

Y a …(i)
A m1
…(ii)
y = m1 ( x + a) + …(iii)

X¢ X and to the parabola y2 = 4b(x + b) is
O B
y = m2 (x + b) + b
m2

C Since two tangents are perpendicular, so

Y¢ m1m2 = –1

and B = (2at2, 0)

Parabola 4.43

Thus, y = m2 (x + b) + b fi x2 – 4x – 1 = 0
m2 fi (x - 2)2 = ( 5)2

y = — 1 (x + b) - bm1 …(iv) fi x=2± 5
m1
Thus, y = 4 ± 5
Eliminating m1 from Eqs (i) and (iv), we get

m1 ( x + a) + a = - 1 (x + b) - bm1 Therefore, Q = (2 + 5, 4 + 5)
m1 m1
and R = (2 - 5, 4 - 5)
1
fi m1 ( x + a + b) = - m1 (x + a + b) Thus, the mid-point of Q and R is (2, 4).

fi (x + a + b) Ê m1 + 1ˆ = 0 11. Clearly, both the lines pass through (–a, –b) which a
ËÁ m1 ˜¯
point lying on the directrix of the parabola.
fi (x + a + b) = 0 ( m1 π 0)
fi (x + a + b) = 0 Thus, m1m2 = –1, since tangents drawn from any point
on the directrix are always mutually perpendicular.
Hence, the result. 12. The equation of any tangent to the parabola y2 = 4x is

8. Let the focal chord be y = mx + c which is passing y = mx + 1
m

through the focus. So fi my = m2x + 1

0 = 4m + c fi m2x – my + 1 = 0 …(i)

fi c = –4m Also (i) is a tangent to the circle (x – 3)2 + y2 = 9

Thus, the focal chord is y = mx – 4m Thus, the length of the perpendicular from the centre to

mx – y – 4m = 0 …(i) the tangent is equal to the radius of the circle.
(i) is tangent to the circle (x – 6)2 + y2 = 2, so
So, 3m2 + 1 = 3
6m - 4m = 2 m4 + m2
m2 + 1
fi (3m2 + 1)2 = 9(m4 + m2)
fi 2m = 2 fi 9m4 + 6m2 + 1 = 9(m4 + m2)
m2 + 1 fi 3m2 = 1
fi m=± 1
fi 4m2 = 2(m2 + 1)
3
fi 2m2 = 2
fi m2 = 1 Hence, the equation of the common tangent which lies

fi m = ±1 above x-axis is y = x + 3
3
9. The co-ordinates of the latus rectum are

L = (a, 2a) = (1, 2) and L¢ = (a, –2a) = (1, –2) fi x - 3y + 3 = 0
13. The equation of any tangent to the parabola y2 = 8x is
The equation of the tangent at L is

yy1 = 2(x + x1) y = mx + 2 …(i)
fi 2y = 2(x + 1) m

fi y = (x + 1) …(i)

The equation of the tangent at L is Since (i) is also a tangent of xy = –1, so

–2y = 2(x + 1)

fi y = –(x + 1) …(ii) x ÊÁË mx + 2 ¯ˆ˜ = -1
m
Solving Eqs (i) and (ii), we get

x = –1, y = 0 mx2 + 2x + 1 = 0
m
Hence, the point of intersection is (–1, 0). fi

10. The equation of the tangent at P to the parabola fi m2x2 + 2x + m = 0
y2 = 8x is
Since it will provide us equal roots, so
4y = 4(x + 2)
D=0
fi y = (x + 2) …(i) fi 4 – 4m3 = 0
Given parabola is y2 = 8x + 5 …(ii) fi m3 = 1

Solving Eqs (i) and (ii), we get fi m=1

(x + 2)2 = 8x + 5 Hence, the equation of the common tangent is y = x + 2.
fi x2 + 4x + 4 = 8x + 5

4.44 Coordinate Geometry Booster

14. Given parabola is y = x2. fi x = - p ± p = p , - 3p
So, 4a = 1 2 22
a = 1/4
The equation of any tangent is when x= p , then, y = ± p
2
x = my + a = my + 1
m 4m Thus the point of intersection are

which is also a tangent of ÊÁË p, pˆ¯˜ and ÊÁË p , - p˜ˆ¯
2 2
y = –x2 + 4x – 4
16. Let the parabola be y2 = 4ax and the two points on the
fi x = m(- x2 + 4x - 4) + 1 parabola are
4m

fi 4mx = 4m2(–x2 + 4x – 4) + 1 P(at12, 2at1) and Q(at22, 2at2 )

fi –4m2x2 + 4(4m2 – m) + (1 – 16m2) = 0 P
fi 4m2x2 – 4(4m2 – m)x – (1 – 16m2) = 0

Since it will provide us equal roots, so

D=0

fi 16(4m2 – m)2 + 16m2(1 – 16m2) = 0
fi (4m2 – m)2 + m2(1 – 16m2) = 0
fi 16m4 – 8m3 + m2 + m2 – 16m4 = 0

fi –8m3 + 2m2 = 0 Q
fi 4m3 – m2 = 0

fi m = 0, 1 The equation of the normal to the parabola at
4
P(at12, 2at1) is
Hence, the equation of the common tangents are

y = 0 and y = 4(x – 1) y = - t1x + 2at1 + at13
15. Given parabola is y2 = 2px. which meets the parabola again at Q(at22, 2at2 ) )

So, the focus is ÁËÊ p , 0˜ˆ¯ . Thus, 2at2 = - at1t22 + 2at1 + at13
2

Y fi 2a(t2 - t1) + at1(t22 - t12 ) = 0

X¢ fi (t2 – t1)[2a + at1(t2 + t1)] = 0

fi [2 + t1(t2 + t1)] = 0
fi t12 + t1t2 + 2 = 0

O C(p/2, 0) X Hence, the result.
17. Given parabola is y2 = x.

x = p/2 So, 4a = 1
Y¢
fi a= 1
4

Clearly, the equation of the circle is The equation of any normal to the parabola

ÁËÊ x p ˆ˜¯ 2 y = mx - 1 m - 1 m3
2 24
— + y2 = p2
which is passing through (c, 0). So
ËÊÁ x p ¯˜ˆ 2
2 mc - 1 m - 1 m3 = 0
fi — + 2px = p2 24

fi x2 - px + p2 + 2px = p2 fi 4mc – 2m – m3 = 0
4 fi m3 + 2(1 – 2c) = 0
fi m2 + 2(1 – 2c) = 0
fi x2 + px + p2 = p2 fi (1 - 2c) = — m2
4
2
2
ÊÁË x p ˆ¯˜ fi (1 - 2c) = — m2 < 0
fi + 2 = p2 2

fi ÁËÊ x + p ˆ˜¯ = ±p fi c>1
2 2

Parabola 4.45

18. The equation of any tangent to the parabola y2 = 8x is 20. Prove that from any point P(at2, 2at) on the parabola
y = mx + 2 y2 = 4ax, two normals can be drawn and their feet
m
Q and R have the parameters satisfying the equation
Clearly, l2 + lt + 2 = 0.
tan (± 45°) = m - 3 21. Given parabola is x2 = 8y.
1 + 3m
We have 4a = 8

fi m - 3 = ±1 fi a=2
1 + 3m The equation of the normal to the parabola x2 = 8y at

fi m - 3 = 1, m - 3 = -1 (- 2am1, am12 ), (- 2am2, am22 ) are
1 + 3m 1 + 3m
x = m1y - 2am1 - am13 …(ii)
fi m = - 2, 1
2 and x = m2 y - 2am2 - am23 …(i)

Hence, the equations of tangents are Let (h, k) be the point of intersection.
y = - 2x - 1, y = x + 4
2 Thus, h = –a(m1 + m2) …(iii)
and k = 2a + a(m12 + m22 - 1) …(iv)
fi y = –2x – 1, x – 2y + 8 = 0
Solving y2 = 8x and y = –2x – 1 we get, the point of From Eqs (iii) and (iv), we get

x2 = y - 2a -1= y — 2a — a = y - 3a
a2 a a a
Ê 1 2ˆ¯˜
intersection is ËÁ 2 , - . fi x2 = a(y – 3a)
fi x2 = 2(y – 6)
Again, solving y2 = 8x and x – 2y + 8 = 0 we get the
which is the required locus.
point of intersection is (8, 8).
22. The equation of the tangent at P(at12, 2at1) to the
Thus, the point of contacts are Ê 1 , - 2ˆ¯˜ and (8, 8). parabola y2 = 4ax is
ËÁ 2
yt1 = x + at12
19. Clearly, the ends of a latus rectum are L(a, 2a) and
L¢ = (a, –2a). fi x - yt1 + at12 = 0

Y¢ Q¢ fi at12 - yt1 + x = 0 …(i)
L
Also, the equation of the tangent at Q(2bt2, bt22 ) to the

X¢ O X parabola x2 = 4by is

xt2 = y + bt22 …(ii)

L¢ Q It is given that the tangents (i) and (ii) are perpendicu-
Y¢
lar, so

The equation of the normal at L is Ê 1 ˆ ◊ t2 = -1
y = mx – 2am – am3 ËÁ t1 ˜¯

putting m = –1, we get fi t2 = –t1
Equation (ii) reduces to
y = –x + 2a + a
fi x + y = 3a -xt1 = y + bt12

The equation of the normal at L is fi xt1 + y + bt12 = 0
y = mx – 2am – am3
fi bt12 + xt1 + y = 0 …(iii)
putting m = 1, we get
Solving Eqs (i) and (iii), we get
y = x – 2a – a
fi x – y = 3a t12 y2) = bx t1 ay = ax 1 by
On solving x + y = 3a and y2 = 4ax, we get -(x2 + - +

Q = (9a, –6a) Eliminating t1, we get
Again, solving x – y = 3a and y2 = 4ax, we get (ax + by)(x2 + y2) + (bx – ay)2 = 0

Q¢ = (9a, –6a) which is the required locus of the point of intersection
of two tangents.
Thus, the length of QQ¢ = (9a — 9a)2 + (6a + 6a)2

= 12a

Hence, the result.

4.46 Coordinate Geometry Booster

23. Let P, Q, and R be three points on the parabola. fi 9b2 = 4a(a – 8a)
Hence, the locus of G(a, b) is 9y2 = 4a(x – 8a)
Y 24. Let P = (at12, 2at1) and Q = (at22, 2at2 ) .
P(t1)
Y
X¢ O X P
Q(t2)
X¢ O M(h, k) X

R(t3)
Y¢

Let the co-ordinates of the centroid be G(a, b). Q
Clearly, a = a(t12 + t22 + t32 ) Y¢

3 Slope of PQ = 2a(t2 - t1) = 2
a(t22 - t12 ) (t1 + t2 )
and b = 2a(t1 + t2 + t3)
3 Equation of PQ is

Now y - 2at1 = (t1 2 (x - at12 )
+ t2)
2a(t1 - t2 ) 2
Slope of PQ = a(t12 - t22 ) = (t1 + t2 ) fi 2x – (t1 + t2)y = –2at1t2 …(i)
…(ii)
and the slope of PR = 2a(t1 - t3) = 2 Now, m1 = m(OP) = 2
a(t12 - t32 ) (t1 + t3) t1

Clearly, –QPR = 60° 2
t2
2-2 and m2 = m(OQ) =

Thus, tan (60°) = t1 + t2 t1 + t3 As OP is perpendicular to OQ, so
1+ 2 ◊2
m1m2 = –1
t1 + t2 t1 + t3 fi 4 = -1

fi 3 = 2(t3 - t2 ) t1t2
(t1 + t2 )(t2 + t3) + 4
fi t1t2 = –4
3 [(t1 + t2 )(t2 + t3) + 4] = 2(t2 - t3) …(i) From Eqs (i) and (ii), we get
Similarly, –Q = 60° and –R = 60°
2(x – 4a) – (t1 + t2) y = 0
Let M(h, k) be the mid-point of PQ. So

Thus, 3 [(t1 + t2 )(t1 + t3 ) + 4] = 2(t3 - t1) …(ii) h = a (t12 + t22 ) and k = a (2t1 + 2t2 )
2 2
and 3 [(t1 + t3 )(t2 + t3 ) + 4] = 2(t1 - t2 ) …(iii)
Adding Eqs (i), (ii) and (iii), we get a
fi h = 2 (t12 + t22 ) and k = a(t1 + t2 )

3(t1t2 + t2t3 + t3t1) + (t12 + t22 + t32 + 12) = 0 Now, k2 = a2(t1 + t2)2
fi k 2 = a2{(t12 + t22 ) + 2t1t2}
fi 3(t1t2 + t2t3 + t3t1) + 3a + 12 = 0
a
fi k2 = a2(2h + 2(–4))
fi (t1t2 + t2t3 + t3t1) + a + 4 = 0
a Hence, the locus of M(h, k) is
y2 = 2a2(x – 4)
fi a(t1t2 + t2t3 + t3t1) + a + 4a = 0
fi 2a(t1t2 + t2t3 + t3t1) + 2a + 8a = 0 25. Let AB be a chord of a parabola, in which
fi a{(t1 + t2 + t3)2 - (t12 + t22 + t32 )} + 2a + 8a = 0
A = (t12, 2t1), B = (t22, 2t2 )

Ï 9b 2 3a ¸ Slope of AB = 2
Ì 4a 2 a ˝
fi a Ó - ˛ + 2a + 8a = 0 fi 2 =2
t1 + t2
fi 9b 2 - 3a + 2a + 8a = 0
4a fi t1 + t2 = 1

Parabola 4.47

Y Therefore, the points P, Q are ÁËÊ - a , a2 ˆ˜¯ , Ê - a , - a ˆ
A 2 ËÁ 2 2 ˜¯

P(h, k) and R, S are (a, 2a) and (a, –2a) respectively.
Thus, the area of the equadrilateral PQRS
X¢ O X
= 1 (PQ + RS) ¥ LM
2

B 1 ÁÊË a a˜ˆ¯ 15a2
2 2 4
Y¢ = ¥ (a + 4a) ¥ + =

Let P be a point which divides AB internally in the ratio 27. Let the point P be (h, k).
1:2
The equation of any normal to the parabola y2 = 4x is

So, h = 2t12 + t22 and k = 4t1 + 2t2 y = mx – 2m – m3 which is passing through P. So
33
k = mh – 2m – m3
3h = (2t12 + t22 ) and 3k = (4t1 + 2t2 )
fi m3 + (2 – h)m + k = 0 …(i)

which is a cubic equation in m.

Eliminating t1 and t2, we get Let its roots are m1, m2, m3.

8ˆ 2 Thus, m1 + m2 + m3 = 0
9 ˜¯
Ê k - = 4 ÊËÁ h - 2 ˆ¯˜ m1m2 + m2m3 + m3m1 = (2 - h)
ÁË 9 9
m1m2m3 = - k

Thus, the locus of P(h, k) is It is given that m1m2 = a, so

Ê 8ˆ 2 4 Ê 2ˆ k
ÁË 9¯˜ 9 ÁË 9 ¯˜ a
y - = x - m3 = -

Hence, the vertex is Ê 2 , 8ˆ . Since m3 is root of Eq. (i), so
ÁË 9 9¯˜
m33 + (2 - h)m3 + k = 0

26. The equation of any tangent to the parabola can be con- fi — k3 + (2 - h) Ê - k ˆ + k = 0
sidered as a3 ÁË a ˜¯
y = mx + a
m fi — k2 + (2 - h) Ê - 1 ˆ +1= 0
a3 ËÁ a ˜¯
Y
R(a, 2a) fi –k2 – (2 – h)a2 + a3 = 0
fi k2 + (2 – h)a2 – a3 = 0
P(–a/2, a/2)
Hence, the locus of P is
X¢ O M X y2 + (2 – x)a2 – a3 = 0
Q(–a/2, –a/2)
which represents a parabola.
S(a, –2a)
Y¢ clearly, a = 2, since a = 2 satisfies the given parabola
y2 = 4x.
i.e. m2x – my + a = 0
28. Given parabola is
As we know that the length of perpendicular from the y2 – 2y – 4x + 5 = 0

centre to the tangent to the circle is equal to the radius fi (y – 1)2 = 4x – 4 = 4(x – 1)
fi Y 2 = 4X
of a circle.
where X = (x – 1), Y = (y – 1)
Thus, a =a
m4 + m2 2 So, the directrix is

fi m4 + m2 = 2 X+a=0
fi m4 + m2 – 2 = 0 fi (x – 1) + 1 = 0
fi (m2 + 2)(m2 – 1) = 0 fi x=0

fi m = ±1 Any point on the parabola is
P(1 + t2, 2t + 1)
Hence, the equation of the tangents are
The equation of the tangent at P is
t(y – 1) = x – 1 + t2

which meets the directrix x = 0 at

y = x + a, y = –x – a

4.48 Coordinate Geometry Booster

Q ËÁÊ 0, 1 + t - 1t ˜¯ˆ (am2 – 1, 4 – 2am)
i.e. (4m2 – 1, 4 – 8m)
Let the co-ordinates of R be (h, k).
Since it divides QP externally in the ratio 1 :1 , so Q is which is passing through (14, 7). So
3 = 15m – 8m – 4m3
2
the mid-point of R and P. fi 4m3 – 7m + 3 = 0
fi 4m3 – 4m2 + 4m2 – 4m – 3m + 3 = 0
\ h + 1 + t2 = 0 and 1 + t - 1 = k + 1 + 2t fi 4m2(m – 1) + 4m(m – 1) – 3(m – 1) = 0
2 t2 fi (m – 1)(4m2 + 4m – 3) = 0
fi (m – 1) = 0, (4m2 + 4m – 3) = 0
fi t2 = - (h + 1) and t = 2 fi (m – 1) = 0, (4m2 + 6m – 2m – 3) = 0
1- k
fi (m – 1) = 0, (2m – 1)(2m + 3) = 0
Thus, (k 4 + (h + 1) = 0
- 1)2 fi (m – 1) = 0, (2m – 1) = 0, (2m + 3) = 0

fi (k – 1)2(h + 1) + 4 = 0 fi m = 1, 1 , - 3
22

Hence, the locus of R(h, k) is Hence, the feet of the normals are
(y – 1)2(x + 1) + 4 = 0
(3, –4), (0, 0), (8, 16)

29. Now, f (x + 1) = - (x + 1)2 + x + 2 31. Let the parabola be y2 = 4ax.
2
The equation of any tangent to the given parabola is

= - x2 - 2x - 1 + 2x + 4 = - x2 + 3 y = mx + a …(i)
22 m

f (1 - x) = - (1 - x)2 + 1 - x + 1 Let the fixed point be (h, k).
2
Also, The equation of any line passing through P and perpen-

= -1 + 2x - x2 + 4 - 2x = -x2 + 3 dicular to (i) is
23
y - k = - 1 (x - h) …(ii)
m

Thus, y = - x2 + x + 1 is symmetric about the line Eliminating m between Eqs (i) and (ii), we get
2
y = - (x - h) - a( y - k)
x=1 ( y - k) (x - h)
Also, given curve is y = - x2 + x + 1
which is the required locus.
2
32. Let P(at12, 2at1) and (at22, 2at2 ) be the extremities of
fi 2y = –x2 + 2x + 2 the focal chord.
fi x2 – 2x = –2y + 2
Thus, t1t2 = –1
fi (x - 1)2 = - 2y + 3 = —2 Ê y — 3ˆ Let the points of intersection of the normals at P and Q
ËÁ 2˜¯ are R(a, b).

fi X2 = –2Y Then a = a(t12 + t22 + t1t2 + 2)

X = (x - 1), Y = Ê y — 3ˆ = a(t12 + t22 — 1 + 2)
ËÁ 2¯˜
= a(t12 + t22 + 1)

Axis is X = 0 = a[(t1 + t2)2 – 2t1t2 + 1]
= a[(t1 + t2)2 + 3]
fi x–1=0 …(i)
and b = –at1t2(t1 + t2) …(ii)
fi x=1
= a(t1 + t2)
which is already proved that the given curve is sym-
From Eqs (i) and (ii), we get,
metric about the line x = 1.

Hence, the result. a = a Ê b 2 ˆ
ÁË a 2 + 3¯˜
30. Given parabola is y2 – 16x – 8y = 0
fi y2 – 8y = 16x = Ê b2 ˆ
fi (y – 4)2 = 16(x + 1) ËÁ a + 3a¯˜
fi Y2 = 16X

where Y = y – 4, X = x + 1

The equation of any normal to the parabola fi b2 = a(a – 3a)
Hence, the locus of (a, b) is y2 = a(x – 3a)
Y = mX – 2am – am3 at (am2, –2am)

y – 4 = m(x + 1) – 8m – 4m3 at

Parabola 4.49

LEVEL IV 3.

4. The equation of the tangent to the parabola y2 = 4x at

1. The equation of the chord of contact is (t2, 2t) is yy1 = 2(x + x1) …(i)
yy1 – 2(x + x1) = 0 fi 2t ◊ y = 2(x + t2)
fi t ◊ y = (x + t2)
fi 2y – 2(x – 1) = 0 fi x – ty + t2 = 0
fi y = (x – 1)
and the equation of the normal to the ellipse is
Y
a2x - b2 y = a2 - b2
A x1 y1
fi 5x - 4y = 5 - 4 = 1
P(–1, 2) M X 5 cos j 2 sin j
X¢ O
fi 5x - 2y = 1
B cos j sin j

Y¢ Equations (i) and (ii) are identical, so

On solving y2 = 4x and y = x – 1, we get, the co-ordi- 1 = -t = -t2
nates of the points A and B. 5 -2 1
Thus, A = (3 - 2 2, 2 - 2 2)
cos j sin j
and B = (3 + 2 2, 2 + 2 2)
fi cos j = t sin j = - t2
Now, the length AB = 8 52
Therefore, the area of DPAB
fi cos j = - 5t2, sin j = - 2t
= 1 ¥ PM ¥ AB
2 Now, cos2 j = 5t4
fi 1 – sin2 j = 5t4
= 1 ¥ —1 — 2 — 1 ¥ 8 fi 1 – 4t2 = 5t4
22 fi 5t4 + 4t2 – 1 = 0
fi (t2 + 1)(5t2 – 1) = 0
= 1 ¥ 4 ¥8 fi (5t2 – 1) = 0
22
fi t=± 1
= 8 2 s.u. 5

2. Given curve is y2 – 2x – 2y + 5 = 0 Now, cos j = - 5 ¥ 1 = - 1
fi y2 – 2y = 2x – 5 55
fi (y – 1)2 = 2x – 5 + 1
fi (y – 1)2 = 2x – 4 = 2(x – 2) fi j = cos-1 Ê - 1ˆ
fi Y2 = 2X ÁË 5 ¯˜
where X = x – 2, Y = y – 1
which represents a parabola. 5. The equation of the normal to the parabola
y2 = 4ax is y = mx – 2am – am3
Now, the focus = (a, 0)
fi X = a, Y = 0 which is passing through (5a, 2a).
fi 2a = 5am – 2am – am3
fi x - 2= 1, y -1= 0 fi 2 = 5m – 2m – m3
2 fi m3 – 3m + 2 = 0
fi m3 – m2 + m2 – m – 2m + 2 = 0
fi x= 5, y =1 fi m2(m – 1) + m(m – 1) – 2(m – 1) = 0
2 fi (m – 1)(m2 + m – 2) = 0
fi (m – 1) = 0, (m2 + m – 2) = 0
Hence, the focus is Ê 5 , 1¯ˆ˜ .
ÁË 2 fi (m – 1) = 0, (m – 1)(m + 2) = 0

Also, the directrix: X + a = 0 fi m = 1, –2
fi x-2+ 1=0
Hence, the equation of the normals are
2
fi x=3 y = x – 2a – a = x – 3a

2 and y = –2x + 4a + 8a = –2x + 12a.

4.50 Coordinate Geometry Booster

6. Let the point of intersection be P(h, k). Therefore,
x = 2a + a(t12 + 4t12 + 2t12 ) = 2a + 7at12
The equation of any normal to the given parabola is

x = my – 2am – am3 ( a = 2) fi t12 = ËÁÊ x - 2a ˜ˆ¯
fi x = my – 4m – 2m3 7a

which is passing through P. Ê x - 2a ˆ 1/ 2
fi h = mk – 4m – 2m3 ËÁ 7a ¯˜

fi 2m3 + (4 – k)m + h = 0 fi t1 =

Let its roots be m1, m2, m3. and y = - at1 ◊ 2t1(t1 + 2t1) = - 6at13

m1 + m2 + m3 = 0

m1m2 + m1m3 + m2m3 = 4 - k 3/2
2
fi y = - 6a ÁËÊ x - 2a ˆ¯˜
7a
and m1m2m3 = - h
2 3/2
-18 ÁÊË x - 6ˆ
m1m2m3 = -h fi y = 21 ˜¯
2
Now,

h 18 ÊÁË x - 6ˆ 3/2
2 21 ¯˜
fi m3 = , ( m1m2 = –1) fi y + = 0 .

Also, m1 + m2 = –m3 8. Given circle is x2 + (y – 3)2 = 5.
The equation of any tangent to the parabola y2 = x.
and m1m2 + m1m3 + m2m3 = 4 - k
2 can be considered as

fi —1 + (m1 + m2 )m3 = 4 - k y = mx + 1
2 4m

fi —1 - m32 = 4 - k fi 4my = 4m2x + 1
2 fi 4m2x – 4my + 1 = 0

fi —1 - h2 = 4 - k Now, OM = 5
42
fi 0 - 12m + 1 = 5
fi 4 + h2 = k — 4 16m4 + 16m2
42
fi (1 – 12m)2 = 5(16m4 + 16m2)
fi 4 + h2 = 2k – 8 fi (1 – 12m)2 = 5(16m4 + 16m2)
fi h2 = 2(k – 6) fi 1 – 24m + 144m2 = 80m4 + 80m2
Hence, the locus of P is x2 = 2(y – 6). fi 80m4 – 64m2 + 24m – 1 = 0
7. Let the parabola be y2 = 4ax and two points on the fi 80m4 – 40m3 + 40m3 – 20m2

parabola are P(at12, 2at1) and Q(at22, 2at2 ) ., where –44m2 + 22m + 2m – 1 = 0
a=3 fi 40m3(2m – 1) + 20m2(2m – 1)

It is given that –22m(2m – 1) + 1(2m – 1) = 0
fi (2m – 1)(40m3 + 20m2 – 22m + 1) = 0
2at1 = 1
2at2 2 fi m=1
2
fi t1 = 1
t2 2

fi 2t1 = t2 Hence, the equation of the common tangent be
The equation of the normal at P(at12, 2at1) and
Q(at22, 2at2 ) are y= x+1
22

y = - t1x + 2at1 + at13 …(i) fi 2y = x + 1,

fi x – 2y + 1 = 0

and 9. The equation of the normal to the given parabola
y = - t2x + 2at2 + at23
…(ii) y2 = 8(x – 1) is
y = m(x – 1) – 2am – am3
Solving Eqs (i) and (ii), we get
fi y = m(x – 1) – 4m – 2m3

x = 2a + a(t12 + t22 + t1t2 ) fi y = mx – 5m – 2m3 …(i)
and y = –at1t2(t1 + t2)
Let the co-ordinates of the point of intersection of the

tangents be P(h, k).

Parabola 4.51

Thus, yy1 = 8 Ê x + x1 ˆ -8 Given parabola is
ËÁ 2 ¯˜ y2 = 4x

fi yk = 4(x + h) – 8 = 4(x + h – 2) fi 2y dy = 4
dx
fi y = 4 (x + h — 2)
k fi dy = 2
dx y
fi y = 4 x + 4 (h — 2) …(ii)
kk Also, given circle is
x2 + y2 – 12x + 31 = 0
Comparing Eqs (i) and (ii), we get
fi 2x + 2y dy - 12 = 0
m = 4 , 4 (h — 2) = - (5m + 2m3) dx
kk
fi x + y dy - 6 = 0
fi 4 (h - 2) = - Ê ◊ 4 + 2 ◊ Ê 4ˆ 3ˆ dx
k ËÁ 5 k ËÁ k ˜¯ ¯˜
fi dy = 6 — x
Ê 2.ÊËÁ 4ˆ 2 ˆ dx y
ËÁ k ¯˜ ˜¯
fi (h - 2) = - 5. + Since the tangents are parallel, so their slopes are the
same.
fi (h – 2)k2 = –(5k2 + 32) Thus, 2 = 6 — x

Hence, the locus of P(h, k) is yy
(2 – x)y2 = (5y2 + 32)
fi x=4
10. Let AB be a double ordinate, where When x = 4, then y2 = 16
fi y = ±4
A = (at12, 2at1), B = (at22, 2at2 ) Thus, the point Q is (4, 4).
Therefore, the shortest distance,
Let P(h, k) be the point of trisection. Then
3h = 2at2 + at2 and 3k = 4at – 2at PQ = CQ – CP

fi 3h = 3at2 and 3k = 2at = (6 - 4)2 + (4 - 0)2 - 1
fi h = at2 and 3k = 2at
= 20 - 1
Solving, we get
= (2 5 - 1)
t2 = h and t = 3k
a 2a

Ê 3k ˆ 2 h 13. Let the parabola be y2 = 4ax
ËÁ 2a˜¯ a
fi = The equation of the tangent to the parabola at (a, 2a) is
y ◊ 2a = 2a(x + a)
fi 9k 2 = h
4a2 a fi y=x+a
fi x–y–a=0
fi 9k2 = 4ah
The equation of a circle touching the parabola at (a, 2a)
Hence, the locus of P(h, k) is
9y2 = 4ax is
(x – a)2 + (y – 2a)2 + l(x – y – a) = 0
11. Do yourself.
12. Given circle is x2 + y2 – 12x + 31 = 0 which is passing through (0, 0). So
a2 + 4a2 – la = 0
fi (x – 6)2 + y2 = 5
fi l = 5a
The centre is (6, 0) and the radius is 5.
Thus, the required circle is
x2 + y2 – 7ax + ay = 0

Y Hence, the radius = Ê 7a ˆ 2 + Ê aˆ2 = 50a2 = 5a
Q ÁË 2 ¯˜ ËÁ 2 ¯˜ 42

P 14. The tangent to y2 = 4x in terms of m is

O C(6, 0) X y = mx + 1
m

and the normal to x2 = 4by in terms of m is

y = mx + 2b + b
m2

4.52 Coordinate Geometry Booster

If these are the same line, then Clearly,

1 = 2b + b 4(t1 + t2) = 4
m m2 fi (t1 + t2) = 1

fi 2bm2 – m + b = 0 Now, m = m( AB) = 2 = 2 = 2
t1 + t2 1
For two different tangents, we get

D>0 Hence, the slope of the normal chord is 2.
fi 1 – 8b2 > 0
fi 8b2 < 1 5. Normals to y2 = 4ax and x2 = 4by in terms of m are
y = mx – 2am – am3

fi b2 < 1 and y = mx + 2b + b
8 m2

fi |b| < 1 For a common normal,
22
2b + b = - 2am - am3
m2
which is the required condition.
15. Given parabola is y2 = 8x fi 2bm2 + b + 2am3 + am5 = 0

Extremities of the latus rectum are (2, 4) and (2, –4). fi am5 + 2am3 + 2b2 + b = 0
Since any circle is drawn with any focal chord as its
diameter touches the directrix, the equation of the re- Thus, the number of common normals is 5.
quired circle is 6. We have, x + y = 2(t2 + 1)

(x – 2)(x – 2) + (y – 4)(y + 4) = 0 and x – y = 2t
fi x2 + y2 – 4x – 12 = 0 Eliminating t, we get,

Hence, the radius = 4 + 12 = 4 . Ê Ê x - yˆ 2 ˆ
ÁË ËÁ 2 ¯˜ + 1˜¯
x + y = 2

Integer Type Questions fi x + y = (x - y)2 + 2
2
1. A circle and a parabola can meet at most in four points.
fi 2(x + y) = (x – y)2 + 4
Thus, the maximum number of common chords is fi (x – y)2 – 2(x + y) + 4 = 0
fi (x – y)2 = 2(x + y – 2)
4C2 = 4 ¥ 3 = 6 . Comparing with y2 = 4ax, we get
2
4a = 2
2. Clearly, both the lines pass through (–a, b) which a Thus, the length of latus rectum is 2.
7. Both the given curves are symmetrical about the line
point lying on the directrix of the parabola. y = x.
If the line AB is the shortest distance then at A and B the
Thus, m1m2 = –1, since tangents are drawn from any slopes of the curve should be equal to 1.
point on the directrix always mutually perpendicular. For y2 = x – 1, dy = 1 = 1

Hence, the value of (m1m2 + 4) is 3. dx 2y
3. Let AB be a normal chord, where
fi y=1
A = (at12, 2at1), B = (at22, 2at2 ) 2

Now, the normal at A meets the parabola again at B, so

t2 = - t1 - 2 and t1t2 = –4 Y
t1 y=x

Solving, we get

t12 = 2 B

Thus, m = m( AB) = 2 A
t1 + t2 X

fi m = 2 = - t1 = 2
(- 2/t1)
x= 1 +1= 5
fi (m2 + 3) = 2 + 3 = 5 Then 4 4.
4. Let AB be a normal chord, where
Ê 5 1ˆ Ê 1 5ˆ
A = (at12, 2at1), B = (at22, 2at2 ) Therefore, A = ËÁ 4 , 2 ˜¯ and B = ËÁ 2 , 4 ¯˜

Parabola 4.53

Hence, the shortest distance, x = 1 - (-2 + 2 2) = 3 - 2 2

Ê 5 1ˆ 2 Ê 1 5ˆ 2 when y = - 2 - 2 2 , then
ËÁ 4 2¯˜ ÁË 2 4¯˜
d= - + - x = 1 - (-2 - 2 2) = 3 + 2 2

Ê 3 ˆ 2 Ê 3ˆ2 18 = 3 2 Let the chord be AB,
ÁË 4 ¯˜ ËÁ 4¯˜ 16 4 where A = (3 - 2 2, - 2 + 2 2),
= + =
and B = (3 + 2 2, - 2 - 2 2)
Hence, the value of
(8d2 – 3) Hence, the length of the chord AB
=9–3=6
= (4 2)2 + (4 2)2 = 32 + 32 = 8
8. The equation of any tangent to the parabola y2 = 4x is
11. Given L: y = –x + k
y = mx + a = mx + 1 Here, m = –1 and c = k
mm The line L will be a normal to the parabola, if
c = –2am – am3
which is passing through (2, 3). So fi k = –2 ¥ 3 ¥ (–1) – 3 ¥ (–1)3
fi k=6+3=9
3 = 2m + 1 Hence, the value of k is 9.
m
12. Given parabola is
fi 2m2 – 3m + 1 = 0 y2 – 4x – 2y – 3 = 0
fi 2m2 – 2m – m + 1 = 0
fi y2 – 2y = 4x + 3
fi 2m(m – 1) – 1(m – 1) = 0 fi (y – 1)2 = 4(x + 1)
The equation of any normal to the given parabola is
fi (m – 1)(2m – 1) = 0
(y – 1) = m(x + 1) – 2am – am3
fi m = 1 and 1/2 fi (y – 1) = m(x + 1) – 2m – m3
which is passing through (–2, 1), so
fi m1 = 1, m2 = 1
2 (1 – 1) = m(–2 + 1) – 2m – m3
fi 0 = –m – 2m – m3
Hence, the value of Ê1 + 1 + 2ˆ˜¯ = 1 + 2 + 2 = 5. fi m3 + 3m = 0
ÁË m1 m2
Hence, the number of distinct normals is 3.
9. Let the co-ordinates of the point R be (at32, 2at3) .

The normal at P meets the parabola again at R, so

t3 = - t1 - 2 …(i)
t1

and the normal at Q meets the parabola again at R, so

t3 = - t2 - 2 …(ii) Previous Years’ JEE-Advanced Examinations
t2
1. The equation of the normal to the parabola y2 = 4ax is y
From Eqs (i) and (ii), we get = mx – 2am – am3 which is passing through (h, 0). So

—t1 - 2 = - t2 - 2 mh – 2am – am3 = k
t1 t2 fi am3 + (2a – h)m + k = 0

fi t2 — t1 = 2 - 2 = 2(t2 - t1) Let its roots are m1, m2 and m3.
t1 t2 t1t2
Thus, m1 + m2 + m3 = 0
fi t2 — t1 = 2(t2 - t1) 2a - h
t1t2
m1m2 + m2m3 + m3m1 = a

fi t1t2 = 2 and m1m2m3 = - k
a
Hence, the value of (t1t2 + 3) = 5
For any real values of m1, m2, m3,
10. Solving, we get,
y2 = 4(1 – y) m12 + m22 + m32 > 0
fi (m1 + m2 + m3)2 – 2(m1m2 + m2m3 + m3m1) > 0
fi y2 = 4 – 4y
fi y2 + 4y – 4 = 0 fi 0 - 2 Ê 2a - hˆ > 0
fi (y + 2)2 = 8 ËÁ a ˜¯

fi y = -2± 2 2 fi h – 2a > 0
fi h > 2a
when y = - 2 + 2 2 , then

4.54 Coordinate Geometry Booster

2. Let the point A be (at2, 2at). c= 3
4
The equation of the normal at A is fi
y = –tx + 2at + at3
5. Let P = (t12, 2t) and Q = (t22, 2t2 )
and the equation of the pair of lines OA and OB is
Y
Ê y + tx ˆ
y2 = 4ax ËÁ 2at + at 3 ¯˜ P

fi (t3 + 2t)y2 = 4tx2 + 4xy X¢ O

p X
2
As –AOB = , we get

co-efficient of x2 + co-efficient of y2 = 0 Q

fi –4t + t3 + 2t = 0 Y¢

fi t3 – 2t = 0 The equation of PQ is

fi t(t2 – 2) = 0

fi t = 0, t = ± 2 2x – (t1 + t2)y = –2t1t2 …(i)

Clearly, t π 0, so t = ± 2 Now, m1 = m(OP) = 2
t1
Thus, slope of normal = - t = ± 2 .
2
3. The equation of the normal to the given parabola is and m2 = m(OQ) = t2
x = my – 2am – am3
As OP is perpendicular to OQ, so
fi x = my – 2m – m3 m1m2 = –1

which is passing through (1, 2)

1 = 2m – 2m – m3 fi 4 = -1 …(ii)
fi m3 = –1 t1t2

fi m = –1 fi t1t2 = –4
From Eqs (i) and (ii), we get
Hence, the equation of the normal is

x = –y + 2 + 1 2(x – 4) – (t1 + t2)y = 0
which is passing through a fixed point (4, 0) on the axis
fi x+y=3
4. The equation of the normal to the given parabola is of the parabola.
Let M(h, k) be the mid-point of PQ.
y = mx – 2am – am3
fi y = mx - 2 ◊ 1 m - 1 m3 Thus, h = 1 (t12 + t22 ) and k = 1 (2t1 + 2t2 )
2 2
44

fi y = mx - 1 m - 1 m3 fi h = 1 (t12 + t22 ) and k = (t1 + t2 )
24 2

fi 4y = 4mx – 2m – m3 Now, k2 = (t1 + t2)2

which is passing through (c, 0), so fi k 2 = (t12 + t22 ) + 2t1t2
0 = 4mc – 2m – m3
fi k2 = 2h + 2(–4)
fi m3 + 2m – 4mc = 0
fi m3 + 2(1 – 2c)m = 0 Hence, the locus of M(h, k) is
fi m = 0, m2 + 2(1 – 2c) = 0 y2 = 2(x – 4) = 2x – 8
fi m = 0, m2 = 2(2c – 1)
6. Given parabola is y2 = 2px. So

fi m = 0, m = ± 2(2c — 1) the focus is ËÊÁ p , 0ˆ¯˜ .
2
So, one normal is always the x-axis.
Y
Let m1 = 2(2c — 1) and m2 = - 2(2c — 1)
X¢ O C(p/2, 0) X
It is given that,
m1m2 = –1 x = p/2
Y¢
fi 2(2c – 1) = 1
fi (2c - 1) = 1

2

fi 2c = 1 + 1 = 3
22

Parabola 4.55

Clearly, the equation of the circle is ÁËÊ 89ˆ˜¯ 2 4 ÊËÁ 2 ˜ˆ¯
9 9
ËÊÁ x p ˆ˜¯ 2 y - = x -
2
— + y2 = p2

Ê p ˆ 2 Hence, the vertex is ÁÊË 2 , 8 ¯ˆ˜ .
ËÁ 2 ˜¯ 9 9
fi x — + 2px = p2

8. Let the three points of the parabola be

fi x2 - px + p2 + 2px = p2 P(at12 , 2at1), Q(at22 , 2at2 ) and R(at32 , 2at3 ),
4
and the points of intersections of the tangents at these
fi x2 + px + p2 = p2
4 points are A[t2t3, a(t2 + t3)], B[t1t3, a(t1 + t3)] and A[t1t2,
a(t1 + t2)]
ËÁÊ x p ˜¯ˆ 2 Now,
2
fi + = p2 at12 2at1 1

ÊÁË x + p ¯ˆ˜ ar(DPQR) = 1 at22 2at2 1
2 2
fi = ±p at32 2at3 1

fi x = - p ± p = p , - 3p = a2(t1 – t2)(t1 – t3)(t3 – t1)
2 22
Also,

when x = p , then, y = ± p ar(DABC) = 1 at2t3 a(t2 + t3 ) 1
2 2 at3t1 a(t3 + t1) 1
a(t1 + t2 ) 1
Thus the point of intersection are at1t2

Ê p, pˆ˜¯ Ê p,- p˜ˆ¯ = 1 a2 (t1 - t2 )(t2 - t3 )(t3 - t1 )
ËÁ 2 ËÁ 2 2
and
\ DPQR = 2
7. Let AB be a chord of a parabola, in which DABC

A = (t12, 2t1), B = (t22, 2t2 ) 9.

10. The equation of the normal to the parabola y2 = 12x

Slope of AB = 2 is y = mx – 2am – am3

fi 2 =2 fi y = mx – 6m – 3m3 …(i)
t1 + t2
Given normal is y = –x + k …(ii)

fi t1 + t2 = 1 Equations (i) and (ii) are identical, so

Y m = –1
A and k = –6m – 3m3 = 6 + 3 = 9

Hence, the value of k is 9.

11 Given parabola is y2 = kx – 8

P(h, k) ÊÁË 8 ˆ˜¯
k
X¢ O X fi y2 = k x -

B Here, 4a = k
fi a=k
Y¢
4
Let P be a point which divides AB internally in the ratio
1 : 2. So So, the directrix is Ê x - 8ˆ + k =0
ÁË k ˜¯ 4

h = 2t12 + t22 and k = 4t1 + 2t2 Given directrix is x – 1 = 0.
33
Thus, 8 - k = 1
fi 3h = (2t12 + t22 ) and 3k = (4t1 + 2t2 ) k4

Eliminating t1 and t2, we get fi 32 – k2 = 4k
fi k2 + 4k – 32 = 0
ÁÊË k 8 ˆ¯˜ 2 4 ÊËÁ h 2 ˜ˆ¯ fi (k + 8)(k – 4) = 0
9 9 9 fi k = 4, –8
- = -

Thus, the locus of P(h, k) is

4.56 Coordinate Geometry Booster

12. Given parabola is If it is a tangent to the curve xy = –1, then

y2 + 4y + 4x + 2 = 0 x ÁËÊ mx + 2 ˜ˆ¯ = -1
m
fi (y + 2)2 = –4x – 2 + 4

fi ( y + 2)2 = - 4x + 2 = — 4 Ê x — 1ˆ fi m2x2 + 2x + m = 0
ÁË 2¯˜
It has equal roots. So,

So, the directrix is x- 1 = a=1 D=0
2 fi 4 – 4m3 = 0
fi m3 = 1
fi x=3
2 fi m =1

Hence, the equation of the common tangent is y = x + 2.

13. Any tangent to the parabola y2 = 4x is 16. Ans. (a)
For the parabola, y2 = 16x , focus = (4, 0)

y = mx + a Let m be the slope of the focal chord.
m
So, its equation is y = m(x – 4) ...(i)

fi y = mx + 1 which is a tangent to the circle
m (x – 6)2 + y2 = 2

fi m2x – my + 1 = 0 …(i) where centre = (6, 0) and radius = 2 .

If it is a tangent to the circle x2 + (y – 3)2 = 9 the length Length of perpendicular from (6, 0) to (i) is equal to r

of the perpendicular from the centre to the tangent is 6m - 4m = 2
m2 + 1
equal to the radius of the circle. So

3m2 + 1 = 3 2m = 2
m4 + m2 m2 + 1

fi (3m2 + 1)2 = 9(m4 + m2) 4m2 = 2(m2 + 1)
fi (9m4 + 6m2 + 1) = 9(m4 + m2) 2m2 = (m2 + 1)
fi 3m2 = 1 m2 = 1

fi m = ± Ê 1ˆ m = ±1
ËÁ 3 ¯˜
17 Given that
Since, the tangent touches the parabola above x-axis, C1: x2 = y – 1
so it will make an acute angle with x-axis, so that m is C2: y2 = x – 1
positive.
Let P(x1, x12 + 1) on C1 and Q( y22 + 1, y2 ) on C2.
Thus m = 1
3

Hence, the common tangent is x - 3y + 3 = 0 .

14. Ans. (c)

If (h, k) be the mid point of line joining the focus (a,

0) and Q(at2, 2at) on the parabola, then h = a + at2 ,
2
k = at. Now, the reflection of the point P in the line y = x can

Eliminating ‘t’, we get, be obtained by interchanging the values of abscissa and

2h = a + a Ê k 2 ˆ the ordinate.
ÁË a 2 ˜¯
Thus, the reflection of the point P(x1, x12 + 1) is
P1(x12 + 1, x1)
fi k2 = 2a ÁÊË h - a ¯ˆ˜ and the reflection of the point Q( y22 + 1, y2 ) is
2
Q1( y2 , y22 + 1)
Now, directrix: ËÊÁ x - a2 ¯ˆ˜ = -a
2 It can be seen clearly that, P1 lies on C2 and Q1 on C1

fi x = 0. Now, PP1 and QQ1 both are perpendicular to the mirror

15. Let the equation of the tangent to the parabola y2 = 8x is line y = x.

y = mx + 2 …(i) Also, M is the mid point of PP1
m
Thus, PM = 1 PP1
2

Parabola 4.57

In triangle PML, PL > PM and m1m2m3 = –k
It is given that m1m2 = a. So
PL > 1 PP1 ...(i)
2 ...(ii) k
m3 = - a
1
Similarly, LQ > 2 QQ1 Since m3 is the roots of (i), so

Adding (i) and (ii), we get, m33 + (2 - h) m3 + k = 0

PL + LQ > 1 (PP1 + QQ1) fi Ê — k ˆ3 + (2 - h) Ê - k ˆ + k = 0
2 ËÁ a ˜¯ ÁË a ¯˜

PQ > 1 (PP1 + QQ1) Ê kˆ3 Ê k ˆ
2 ÁË a˜¯ ËÁ a ˜¯
fi - - (2 - h) + k = 0

PQ is more than the mean of PP1 and QQ1 fi k2 + (2 - h) -1= 0
PQ ≥ min (PP1, QQ1) a3 a

Let min (PP1, QQ1) = PP1

then PQ2 ≥ PP12 fi k2 + (2 – h)a2 – a3 = 0
= (x12 + 1 - x1)2 + (x12 + 1 - x1)2
= 2(x12 + 1 - x1)2 = f (x1) Hence, the locus of P(h, k) is
y2 + (2 – x)a2 – a3 = 0
Now, f ¢(x1) = 4(x12 + 1 - x1)(2x1 - 1)
As this locus is a part of the parabola y2 = 4x so, a2 = 4
Ê ÁÊË 1 ˜ˆ¯ 2 3ˆ and –2a2 + a3 = 0
4 ÁË 2 4¯˜ Thus, a = 2.
= x1 - + (2x1 - 1)
19. The equation of any tangent to the given parabola can
f ¢( x1 ) = 0 gives x1 = 1
2 be considered as

y = mx + a = mx + 1
mm

Also, f ¢(x1) < 0 if x1 < 1 which is passing through (1, 4). So
2 4=m+ 1
m

and f ¢(x1) > 0 if x1 > 1 fi m2 – 4m + 1 = 0
2
Let its roots are m1, m2.
Thus, f(x1) is minimum when x1 = 1 \ m1 + m2 = 4 and m1m2 = 1
2 Let q be the angle between them. Then

Thus, if at x1 = 1 at P is P0 on C1 tan q = m2 - m1
2 1 + m1m2

Ê 1 ËÊÁ 1 ˜¯ˆ 2 ˆ ÁÊË 1, 5 ˆ¯˜ = (m2 + m1)2 - 4m1m2
ËÁ 2 2 + 1¯˜ 2 4 1 + m1m2
So, P0 = , =

Similarly Q0 on C2 will be image of P0 with respect to = 16 — 4 = 2 3
the line y = x 1+1 2

So, Q0 = ÊËÁ 5 , 12¯˜ˆ
4
= 3=p
18. Let the point P be (h, k). 3

The equation of any normal to the given parabola is fi q=p
3
y = mx – 2am – am3
fi y = mx – 2m – m3, since a = 1 20. Given parabola is
y2 – 2y – 4x + 5 = 0
which is passing through P. So …(i)
k = mh – 2m – m3 fi (y – 1)2 = 4x – 4 = 4(x – 1)
fi Y2 = 4X
fi m3 + (2 – h)m + k = 0 where X = (x – 1), Y = (y – 1)
So, the directrix is
Let its roots be m1, m2, m3.
So, m1 + m2 + m3 = 0 X+a=0

m1m2 + m2m3 + m3m1 = (2 – h)

4.58 Coordinate Geometry Booster

fi (x – 1) + 1 = 0 Since it has equal roots, so
fi x=0 fi D=0
Any point on the parabola is
4(2 – t)2 – 4(t2 – 4) = 0
P(1 + t2, 2t + 1) fi (2 – t)2 – (t2 – 4) = 0
The equation of the tangent at P is fi t = 2, 0
Hence, the equation of the common tangent is
t(y – 1) = x – 1 + t2
which meets the directrix x = 0 at y = 4x – 4, y = 0
23.
Q ÊËÁ 0, 1 + t - 1t ¯ˆ˜
Y
Let the co-ordinates of R be (h, k).
Since it divides QP externally in the ratio 1 :1, so Q is P
Q
2
the mid-point of R and P. Thus X¢ O X
M(3, 0)

h + 1 + t2 = 0 and 1 + t - 1 = k + 1 + 2t R
2 t2 Y¢

fi t2 = - (h + 1) and t = 2 Equation of any normal to the given parabola is
1- k
y = mx – 2am – am3 …(i)

Thus, 4 + (h + 1) = 0 Let P = (am12, - 2am1), Q = (am22, - 2am2 )
- 1)2
(k and R = (am32, - 2am3)

fi (k – 1)2(h + 1) + 4 = 0 Equation (i) passing through (3, 0)

Hence, the locus of R(h, k) is So, 0 = 3m – 2am – am3 ( a = 1)
m3 – m = 0,
(y – 1)2(x + 1) + 4 = 0
m(m + 1)(m – 1) = 0

21. Clearly, the vertex is (1, 1) and the focus is S(2, 2) and m = –1, 0, 1
the directrix is x + y = 0.
Let P be (x, y) Thus, m1 = –1, m2 = 0, m3 = 1
Now, SP = PM
fi SP2 = PM2 Now, P = (m12, 2m1) = (1, - 2)

Q = (m22, 2m2 ) = (0, 0)

Ê x +y ˆ 2 and R = (m32, 2m3) = (1, 2)
ÁË 2 ¯˜
fi ( x - 2)2 + ( y - 2)2 = (i) Area of DPQR

fi 2[(x – 2)2 + (y – 2)2] = (x + y)2 1 -2 1
fi 2(x2 + y2 – 4x – 4y + 8) = x2 + y2 + 2xy = 1 0 0 1 = 1 ¥ 4¥1- 2
fi (x2 + y2 – 2xy) = 8(x + y + 2)
fi (x – y)2 = 8(x + y + 2) 22
22. Any point on the parabola y = x2 is (t, t2). 1 21

Now tangent at (t, t2) is (ii) Radius of circumcircle of DPQR = 2
(iii) Centroid of DPQR

xx1 = 1 (y + y1) = Ê m12 + m22 + m32 , 2(m1 + m2 + m3 ) ˆ
2 ÁË 3 3 ˜¯

fi tx = 1 ( y + t2 ) = Ê 2 , 0ˆ˜¯
2 ËÁ 3

fi 2tx – y – t2 = 0 (iv) Clearly, DPQR is a right-angled triangle and right
If it is a tangent to the parabola, y = –(x – 2)2, then angle at Q.
Thus, Circumcentre of DPQR
2tx – t2 = –(x – 2)2 = Mid-point of the hypotenuse PR
= (1, 0).
fi 2tx – t2 = – x2 + 4x – 4

fi x2 + 2(2 – t)x + (t2 – 4) = 0

Parabola 26. Given ellipse is 4.59
24. x2 + 4y2 = 4
X
Y fi x2 + y2 = 1
P(1, –2 2) 41

L¢ Y

N(–30) L
X¢ S(–1, 0) O
X
M(1, 0) R(9, 0)

X¢ S¢ S
V
Q(1, –2 2)

Y¢ QP
Y¢

(i) Co-ordinates of P and Q are

P = (1, 2 2), Q = (1, - 2 2) Thus, e = 1- b2 = 1- 1 = 3
a2 4 2
Area of DPQR = 1 ¥ 4 2 ¥ 8 = 16 2
2 Foci: S = (ae, 0) = ( 3, 0)

Area of DPQS = 1 ¥ 4 2 ¥ 2 = 4 2 and S ¢ = (— ae, 0) = (- 3, 0)
2
End-points of latus recta:
ar (DPQS) = 4 2 =1
Thus, ar (DPQR) 16 2 4 and L = Ê b2 ˆ = ËÁÊ 3, 1 ˜ˆ¯
ËÁ ae, a ¯˜ 2

(ii) R = abc = 23 ¥6 2 ¥ 10 = 3 3 L¢ = Ê —ae, b2 ˆ = ËÁÊ - 3, 1 ˆ¯˜
4D ¥ 10 ¥ 2 2˜¯ˆ ÁË a ¯˜ 2
4 ◊ Ê 1
ÁË 2
Thus, P = ÁËÊ 1 ˆ˜¯ ËÁÊ - 1 ˆ˜¯
(iii) r = D 3, - 2 and Q = 3, - 2
s
As we know that, the focus is the mid-point of the

1 ¥4 2¥8 P and Q.

= 1 2 Thus, the focus of a parabola is Ê 0, - 1ˆ .
¥ (6 2 ËÁ 2¯˜
+ 6 2+4 2)
2
The length of PQ = 2 3

= 32 2 Now, 4a = 2 3 fi a = 3
16 2 2

=2 Thus, the vertices of a desired parabola

25. Given curve is Ê 1 a˜¯ˆ Ê 1 3ˆ
ËÁ 2 ËÁ 0, 2 2 ¯˜
y = - x2 + x + 1 = 0, - ± = - ±
2
Therefore, two desired parabolas are
1 Ê 1 1ˆ
fi y = - 2 ÁË x2 - 2 x - 2˜¯ fi x2 = ± 4a Ê y - Ê - 1 ± 3ˆˆ
ÁË ËÁ 2 2 ˜¯ ˜¯

fi y = - 1 (x - 1)2 + 3 fi x2 = 2 Ê y + 1 + 3ˆ
22 3 ÁË 2 2 ˜¯

fi ÊÁË y — 32¯˜ˆ = - 1 (x - 1)2 or
2
x2 = -2 Ê y + 1 - 3ˆ
3 ÁË 2 2 ˜¯
which is symmetric about the line x =1.
fi x2 = 2 3y + (3 + 3)
Note: A function f(x) is symmetric about the line x =
1 then, f(1 – x) = f(x + 1) or x2 = - 2 3y + (3 — 3)