4.60 Coordinate Geometry Booster
27. Let the co-ordinates P be (at2, 2at). fi 6 = 9m – 2m – m3
fi m3 – 7m + 6 = 0
Y fi m3 – m2 + m2 – m – 6m + 6 = 0
P fi m2(m – 1) + m(m – 1) – 6(m – 1) = 0
fi (m – 1)(m2 + m – 6) = 0
X¢ T OS X
N fi (m – 1)(m – 2)(m + 3) = 0
Y¢ fi m = 1, 2, –3
The equation of PT is yt = x + at2 Thus, the equation of normal can be
So, T is (–at2, 0).
and the equation of PN is y = –tx + 2at + at3 y = x – 3, y = 2x – 12, y + 3x – 33 = 0.
So, N is (2a + at2, 0).
30.
Y
L(2, 4)
A
Let the centroid be G(h, k). B(0, 2) P(1/2, 2) X
X¢
Thus, h = at2 - at2 + 2a + at2 and k = 2at
33 C
x=2
fi h = 2a + at2 and k = 2at M(2, –4)
33 Y¢
ÁËÊ 3h - 2a ˆ¯˜ Ê 3k ˆ 2 Clearly, D LPM = 2
a ËÁ 2a ˜¯ D ABC
fi =
fi 3ËÊÁ h - 2aˆ = 9k 2 fi D1 = 2
3 ˜¯ 4a D2
fi k2 = 4a Ê h - 2aˆ 31. The parabola is x = 2t2, y = 4t
3 ÁË 3 ˜¯ Solving it with the circle, we get
4t4 + 16t2 – 4t2 – 16t = 0
Hence, the locus of G(h, k) is fi t4 + 3t2 – 4t = 0
fi t(t3 + 3t – 4) = 0
y2 = 4a ÁËÊ x - 2a ˆ¯˜ fi t = 0, 1
3 3 So, the points P and Q are (0, 0) and (2, 4), respec-
tively which are also diametrically opposite points on
So the vertex is ÁÊË 2a , 0¯ˆ˜ and focus is (a, 0). the circle.
3 The focus is S = (2, 0)
28. Let A = (t12, 2t1), B = (t22, 2t2 ) 0 01
The area of DPQS = 1 2 4 1
Y
22 0 1
B
C = 1¥2¥4=4
2
X¢ O A X
M
32. Let P = (at 2 , 2at), Q = Ê a , - 2a ˆ
ÁË t2 t ˜¯
Y¢
Then C = Ê t12 + t22 , t1 + t2 ˆ and R = Ê - a, a Ê t - 1ˆ ˆ
ËÁ 2 ˜¯ ËÁ ÁË t ¯˜ ¯˜
Clearly, |t1 + t2| = r Y
fi (t1 + t2) = ±r P
Now, m( AB) = 2(t2 - t1) = 2 =±2
(t22 - t12 ) (t1 + t2 ) r
R
29. Here a = 1 X¢ X
The equation of normal to the parabola y2 = 4x
is y = mx – 2am – am3 Q
Y¢
fi y = mx – 2m – m3
which is passing through (9, 6).
Parabola 4.61
Since R lies on y = 2x + a, so Now,
a Ê t - 1ˆ = -a 1 0 31
ËÁ t ˜¯ 2
ar(DEFG) = 0 4t 1
fi Ê t - 1ˆ = -1 4t2 8t 1
ËÁ t ¯˜
= 1 [4t2 (3 - 4t)]
Ê 1ˆ 2 Ê 1ˆ 2 2
ËÁ t ˜¯ ËÁ t ¯˜
fi t + = t - + 4 = 1 + 4 = 5 = 2t2(3 – 4t)
= (6t2 – 8t3)
Thus, PQ = a ËÊÁ t + 1t ¯˜ˆ 2 = 5a
fi dA = 12t - 24t2 = 12t(1 - 2t)
33. Here, P = (at 2 , 2at), Q = Êa , - 2a ˆ dt
ËÁ t2 t ˜¯
For maximum or minimum,
dA = 0 gives t = 0, 1/2
dt
Y –+ –
0 1/2
P So, t = 1/2 has a point of local maxima.
Thus, G = (0, 4t) = (0, 2) fi y1 = 2
F = (x0, y0) = (4t2, 8t) = (1, 4) fi y0 = 4
X¢ O q S X Area = 2 ËÊÁ 3 - 1ˆ = 2 ¥ 1 = 1
4 2˜¯ 4 2
Q So, y = mx + 3 passes through (1, 4).
Thus, m = 1.
Y¢ 35. The equation of any tangent to the parabola can be con-
Now, t - 1 = -1 sidered as
t y = mx + a = mx + 2
mm
fi ËÁÊ t + 1t ˆ˜¯ 2 = 1 + 4 = 5 Y R(2, 4)
P(–1, 1)
fi Ê t + 1ˆ = 5 X¢ O X
ÁË t ˜¯ Q(–1, –1)
Ê 2 + 2t ˆ 2 Ê t + 1ˆ
t ˜ ËÁ t ˜¯
tan q = Á = =-2 5 Y¢ S(2, –4)
ËÁ 1 - 4 ˜¯ 3
-3 i.e. m2x – my + 2 = 0
34. The tangent at F(4t2, 8t), is As we know that the length of the perpendicular drawn
yy1 = 8(x + x1)
from the centre to the tangent to the circle is equal to
fi y ◊ 8t = 8(x + 4t2)
fi y ◊ t = (x + 4t2) the radius of a circle.
Y Thus, 2= 2
m4 + m2
E(0, 3)
F X fi m4 + m2 = 2
fi m4 + m2 – 2 = 0
G(0, y1) fi (m2 + 2)(m2 – 1) = 0
X¢ O fi m = ±1
Hence, the equation of the tangents are
y = x + 2, y = –x – 2
Therefore, the points P, Q are (–1, 1), (–1, –1) and R, S
Y¢ are (2, 4) and (2, –4) respectively.
Put x = 0, then y = 4t Thus, the area of the equadrilateral PQRS
Thus, pt G is (0, 4t)
= 1 ¥ (2 + 8) ¥ 3 = 15
2
4.62 Coordinate Geometry Booster
36. Given P = (at2, 2at) fi 2yt = a Ê t 2 + 1 + 2ˆ˜¯
ËÁ t2
Since PQ is a focal chord, so the co-ordinates of Q are
Ê a , - 2a ˆ . fi 2yt = a ÁÊË t + 1t ˆ˜¯ 2
ÁË t2 t ˜¯
Also, R = (ar2, 2ar), S = (as2, 2as)
and K = (2a, 0) a Ê 1ˆ 2 a(t2 + 1)2
2t ËÁ t ˜¯ 2t 3
It is given that, fi y = t + =
m(PK) = m(QR)
2a 38. Image of y = –5 about the line x + y + 4 = 0 is x = 1
0 - 2at = 2ar + t Hence, the required distance AB = 4
2a - at2 ar2 - a
fi t2 39. Equation of normals are
x + y = 3 and x – y = 3
r + 1 Hence, the distance from (3, –2) on both the normals is
r2 -
fi t = t r
- 1
t2 2 t2 Thus, 3 - 2 - 3 = r
2
r+1
t t fi r2 = 2
-
fi t2 2 = Ê 1ˆ Ê 1ˆ 40.
ËÁ t ¯˜ ËÁ t ˜¯
r + r -
fi t2 t 2 = Ê 1 1ˆ
- ÁË - t ¯˜
r
fi Ê r - 1ˆ t = (t 2 - 2)
ËÁ t ¯˜
fi rt – 1 = (t2 – 2)
fi r = Ê t2 -1ˆ = ËÁÊ t - 1t ¯˜ˆ
ËÁ t ¯˜
37. Now, S = (as 2 , 2as) = Ê a , 2a ˆ Clearly, P = (at2, 2at) and Q = Ê 16a , - 8a ˆ
ËÁ t2 t ˜¯ ËÁ t2 t ˜¯
Tangent at P, y ◊ t = x + at2 …(i) Area of the triangle OPQ = 3 2
Tangent at S, yy1 = 2a(x + x1) 1 ◊ OP ◊ OQ = 3 2
2
2a Ê a ˆ
fi y ◊ t = 2a ËÁ x + t2 ˜¯ 1 at t2 + 4 ¥ - 4a 16
2t t2
t2 - 3 2t + 4 = 0 + 4 =3 2
fi y ◊ = t Ê x + a ˆ
ÁË t2 ˜¯
Normal at S, y- 2a = - 1 Ê x - a ˆ P = (at 2, 2at) = Ê t2 , ˆ
t t ÁË t2 ˜¯ ËÁ 2 t¯˜
Ê a ˆ when t = 2, P = (1, 2)
ËÁ t2 ¯˜
fi y ◊ t - 2a = - x - when t = 2 2, P = (4, 2 2)
41. Equation of tangent at P( 2, 1) is
fi y ◊t = - Ê x - a ˆ + 2a …(ii)
ÁË t2 ˜¯ 2x+ y=3
Solving (i) and (ii), we get If centre of C2 at (0, a) and the radius equal to 2 3
fi 2 3= a -3
fi yt - at 2 = a + 2a - yt
t2 3
fi a = –3, 9
fi 2yt = at2 + a + 2a
t2
Parabola 4.63
(a) Q2Q3 = 12 (d) ar(DPQ2Q3)
(b) R2R3 = length of the transverse common tangent
= 1 ¥ 12 ¥ 2 = 6 2
= (Q2Q3)2 - (r1 + r2 )2 2
= (12)2 - (2 3 + 2 3)2
42. Equation of normal of parabola is
=4 6 y + tx = 2t + t3
Normal passes through S(2, 8)
8 + 2t = 2t + t3
t3 = 8
t=2
Hence, P = (4, 4) and SQ = radius = 2
(c) Area of DOR2R3
= 1 ¥ R2 R3 ¥ ^r distance from O to the line
2
=1¥4 6¥ 3=6 2
2
CHAPTER Ellipse 5.1
5 Ellipse
CONCEPT BOOSTER Definition 2
An ellipse is the set of all points in a plane, the sum of whose
1. INTRODUCTION distances from two fixed points is a constant, i.e.
An oval is generally regarded as any ovum (egg)-shaped SP + S¢P = 3a
smooth, convex closed curve. The word convex means any
chord connecting two points of the curve lies completely P
within the curve, and smooth means that the curvature does
not change rapidly at any point. The ellipse is a typical oval, S¢ S
but a very particular one with a shape that is regular and can
be exactly specified. Definition 3
A conic
It has two diameters at right angles that are lines of sym-
metry. It is best to reserve the word ellipse for real ellipses, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
and to call others ovals. A diameter is any chord through the
centre of the ellipse. The diameters that are lines of symme- represents an ellipse if
try are called the major axis (2a), and the minor axis (2b),
where a > b. If a = b, we have the very special ellipse, the (i) D π 0 and
circle, which has enough special properties that it should be (ii) h2 – ab < 0, where
distinguished from an ellipse, though, of course, it has all
the properties of an ellipse in addition to its own remarkable ahg
properties. D= h b f
A vertex of a curve is a point of maximum or minimum gf c
radius of curvature. An ellipse has vertices at the ends of
the major axis (minimum) and at the ends of the minor axis Definition 4
(maximum). Let z, z1 and z3 be three complex numbers such that |z – z1| +
|z – z2| = k, where k > |z1 – z2| and k be a positive real number,
2. MATHEMATICAL DEFINITIONS the locus of z is an ellipse.
Definition 1 Y
It is the locus of a point which moves MP P(z)
in a plane in such a way that its dis-
tance from a fixed point is a constant S (Focus) X¢ A¢ F1(z1) O F2(z2) A X
ratio from a fixed straight line. This Z (Directrix)
ratio is always less than 1. This fixed
point is called the focus and the fixed Y¢
straight line is called the directrix. The
constant ratio is called the eccentricity.
5.2 Coordinate Geometry Booster
3. EQUATION OF AN ELLIPSE fi x2 + y2 = 1, b2 = a2 (1 - e2 ) , …
a2 b2
Y
BP This is the standard equation of an ellipse.
M¢ M 4. PROPERTIES OF AN ELLIPSE
X¢ Z¢ A¢ S¢ C S N A Z X Y
B¢
Y¢ BP
Let S be the focus and ZM be the directrix of the ellipse. Draw M¢ S¢ C S N A M
SZ ^ ZM. Divide SZ internally and externally in the ratio X¢ L¢ A¢ B¢ X
e : 1 (e < 1) and let A and A¢ be the internal and external
point of division. L
Then SA = e AZ …(i) Y¢
and SA¢ = e A¢Z …(ii)
Centre
Clearly A and A¢ will lie on the ellipse. A point inside the ellipse which is the mid-point of the line
segment linking the two foci, i.e. the intersection of the major
Let AA¢ = 2a and take C be the mid-point of AA¢ as origin. and minor axes. Here C = (0, 0).
Thus, CA = CA¢ = a …(iii)
Let P(x, y) be any point on the ellipse referred to CA and Major/minor axis
CB as co-ordinate axes.
The longest and the shortest diameters of an ellipse are
Adding Eqs (i) and (ii), we get known as the major axis and the minor axis respectively. The
length of the major axis is equal to the sum of the two genera-
SA + SA¢ = e(AZ + A¢Z) tor lines.
Here, major axis = AA¢ = 2a,
fi AA¢ = e(CZ – CA + CA¢ + CZ) and minor axis = BB¢ = 2b
fi AA¢ = e(2CZ)
fi 2a = 2eCZ Semi-major/Half the major axis
The distance from the centre to the farthest point on the el-
fi CZ = a lipse is known as the semi-major axis.
e
Semi-minor axis/Half the minor axis
Thus the directrix ZM is x = CZ = a . The distance from the centre to the closest point on the ellipse
e is known as the semi-minor axis.
Again subtracting Eqs (i) from (ii), we get
SA – SA¢ = e(A¢Z – AZ) Directrices
fi (CA¢ + CS) – (CA – CS) = e(AA¢)
fi 2CS = e(AA¢) LM and L¢M¢ are two directrices of the ellipse.
fi 2CS = e(2a)
fi CS = ae Thus LM: x = a and L¢M¢: x = - a .
e e
The distance between two directrices: LL¢ = 2a
Thus the focus is S(CS, 0) = S(ae, 0) e
Now draw PM ^ ZM,
Foci (Focus points)
SP = e The two points that define the ellipse is known as the foci.
PM
Here S = (ae, 0) and S¢ = (–ae, 0)
fi SP2 = e2PM2 Distance between two foci: SS¢ = 2ae
ÊÁË a x¯˜ˆ 2 Perimeter
e The perimeter is the distance around the ellipse, i.e.
fi (x - ae)2 + y2 = e2 -
fi x2 + a2e2 – 2aex + y2 = a2 – 2aex + e2x2 Perimeter = p ¥ È 3 (a + b) + ab ˘ .
ÎÍ 2 ˚˙
fi x2(1 – e2) + y2 = a2(1– e2) It is not easy to calculate.
fi x2 + y2 e2 ) =1 Area
a2 a2 (1 - The number of square units it takes to fill the region inside an
ellipse is called the area of an ellipse.
Ellipse 5.3
x2 y2 Thus the co-ordinates of L and L¢ are
a2 b2
If the equation of ellipse is + = 1, then its area is pab. Ê b2 ˆ Ê b2 ˆ
ÁË a ¯˜ ÁË a ¯˜
ae, and ae, - .
If the equation of ellipse is
Ax2 + Bxy + Cy2 = 1, then its area is 2p . Hence the length of the latus rectum,
4AC - B2 LL¢ = 2b2 .
a
Chord
A line segment linking any two points on an ellipse is known Relation amongst a, b, and e:
as the chord of the ellipse. b2 = a2(1 – e2)
Focal chord Eccentricity (e)
A chord of the ellipse passing through its focus is called the
focal chord. e= 1- b2 = a2 - b2
a2 a2
Focal distances
Let P(x, y) be any point on the ellipse. Auxiliary circle
The circle described on the major axis of an ellipse as diam-
Here, SP = ePM = e ËÊÁ a - xˆ¯˜ =a - ex eter is called an auxilliary circle.
e
S¢P = ePM¢ = e ÊËÁ a + x¯ˆ˜ = a + ex Relation to a circle
e A circle is actually a special case of an ellipse. In an ellipse,
if you make the major and minor axes of the same length, the
Now, SP + S¢P = a – ex + a + ex = 2a = constant. result is a circle, with both foci at the center C.
Thus the sum of the focal distances of a point on the ellipse
is constant. 5. PARAMETRIC EQUATION OF AN ELLIPSE
Notes Y
Focal distances are also known as focal radii of the ellipse.
Vertices P
The vertices of the ellipse are the points where the ellipse
meets its major axis. N
Here, A = (a, 0) and A¢ = (–a, 0) are the vertices of the ellipse.
X¢ f X
CM
Co-vertices Y¢
The co-vertices of the ellipse are the points where the ellipse
meets its minor axis. Let the equation of the ellipse be x2 + y2 =1.
Here, B = (0, b) and B¢ = (0, –b) are the co-vertices of the a2 b2
ellipse.
Double ordinate The equation of its auxilliary circle is x2 + y2 = a2.
It is a chord perpendicular to the major axis and intersects the
curve in two distinct points. Let Q be a point on the auxilliary circle x2 + y2 = a2 such
that QP produced is perpendicular to the x-axis.
Latus rectum Thus P and Q are the corresponding points on the ellipse
and the auxilliary circle.
It is a double ordinate, perpendicular to the major axis and
passes through the foci. Here LSL¢ and L1SL1¢ are two latus Let –QCA = j, where 0 £ j < 2p
recta.
Let Q = (a cos j, a sin j) and P = (a cos j, y).
Length of the LR Since P lies on the ellipse x2 + y2 = 1, so we can write
a2 b2
Let the co-ordinates of L and L¢ be (ae, y1) and (ae, –y1). a2cos2j y2
a2 b2
Since L lies on the ellipse x2 + y2 = 1, so we have + =1
a2 b2
a2e2 y12
a2 + b2 =1 fi y2 = b2 sin2 j
fi y = b sin j
y12 = b2 (1 - e2 ) = b2 Ê b2 ˆ = b4 since P lies in the 1st quadrant.
ËÁ a2 ˜¯ a2
fi Thus the parametric equations of the ellipse are
x = a cos j, y = b sin j.
fi y1 = b2/a
5.4 Coordinate Geometry Booster
Notes Proof B P(r cos q, r sin q)
Any point, say P, on the ellipse can be considered as (a cos
j, b sin j). Since the point is known when f is given, then r
it is often called ‘the point f’ or P(j)..
A¢ S¢ C SA
6. IMPORTANT PROPERTIES RELATED TO CHORD AND
FOCAL CHORD B¢
(i) Equation of the chord joining the points P(j1) and We have, SP = a – ex and S¢P = a + ex.
Q(j2) Thus,
P(f1) SP. S¢P = (a2 – e2x2) = (a2 – a2e2 cos2 q)
= a2 + (b2 – a2) cos2 q
Q(f2) = a2(1 – cos2 q) + b2 cos2 q
= a2 sin2 q + b2 cos2 q
The equation of the chord joining the points P(a cos j,
b sin j1) and Q(a cos j2, b sin j2) is (iv) If PQ be a focal chord and S, S¢ are the foci of an el-
lipse, the perimeter of the triangle described by DS¢PQ
x ÁÊË j1 + j2 ˜¯ˆ y sin ÁÊË j1 + j2 ˆ¯˜ is 4a.
a 2 b 2 Proof
P
cos +
= cos ËÊÁ j1 - j 2 ˆ˜¯ S¢ S
2
(ii) The length of a radius vector from the centre drawn in Q
a given direction
We have, perimeter of the DS¢PQ
B P(r cos q, r sin q) = S¢P + S¢Q + PQ
= (S¢P + S¢Q) + (SP + SQ)
r = (S¢P + SP) + (S¢Q + SQ)
= 2a + 2a = 4a
A¢ C A
(v) The length of the focal chord of an ellipse which makes
B¢ an angle q with the major axis is 2ab2 .
a2sin2q + b2cos2q
As we know that the equation of the ellipse is
Proof
x2 y2
a2 + b2 = 1. P
Put x = r cos q, y = r sin q, we have q
S
r 2 cos2q r 2sin 2q
a2 + b2 = 1.
a2b2 Q
b2cos2q + a2sin2q
fi r2 = Let the chord be PQ, where P = (x1, y1), Q = (x2, y2) and
S(ae, 0) be the focus.
fi r= ab The chord PQ be
b2cos2q + a2sin2q (y – 0) = tan q (x – ae) …(i)
which is the required distance from the centre of the Now PQ = SP + SQ
point P(q) on the ellipse x2 + y2 = 1. = a – ex1 + a – ex2 = 2a = e(x1 + x2)
a2 b2 Let the equation of the ellipse be
(iii) Product of the focal radii of an ellipse from any point x2 + y2 =1 …(ii)
a2 b2
x2 y2
P(q) on the ellipse a2 + b2 = 1 is Now PQ = SP + SQ
a2 sin2 q + b2 cos2 q. = a – ex1 + a – ex2
= 2a – e(x1 + x2)
Ellipse 5.5
From Eqs (i) and (ii), we get ËÊÁ a - b ˜ˆ¯
2 b ¯ˆ˜
x2 + tan2q (x - ae)2 =1 fi cos = e
a2 b2 cos
ÊËÁ a +
2
fi b2x2 + a2 tan2 q(x – ae)2 = a2b2
fi b2x2 + a2 tan2 q(x2 – 2aex + a2e2) = a2b2 cos Ê a - bˆ + cos Ê a + bˆ
ÁË 2 ˜¯ - ÁË 2 ˜¯
fi (b2 + a2 tan2 q)x2 – 2a3e tan2 q x fi = e -1
bˆ bˆ e +1
+ a2(a2e2 tan2 q – b2) = 0 cos Ê a - ¯˜ cos Ê a + ¯˜
ÁË 2 ÁË 2
Let its roots are x1 and x2.
Then x1 + x2 = 2a3e tan2q fi tan ÊÁË a2 ¯˜ˆ tan Ê bˆ = e -1
b2 + a2tan2q ÁË 2 ˜¯ e +1
Therefore, PQ = 2a – e(x1 + x2) Hence, the result.
= 2a - e Ê 2a3e tan 2q ˆ (vii) If a and b are the eccentric angles of the extremities
ËÁ b2 + a 2 tan 2q ¯˜
of a focal chord of the ellipse x2 + y2 =1, then the
eccentricity of the ellipse is a2 b2
Ê b2 + a2tan2q - a2e2tan 2q ˆ
= 2a ËÁ b2 + a2tan2q ¯˜
sin a + sin b
Ê b2 + a2 (1 — e2 ) tan 2q ˆ sin (a + b)
ËÁ b2 + a2 tan 2q ¯˜
= 2a Proof
Ê b2 + b2tan 2q ˆ P
ÁË b2 + a2tan 2q ¯˜
= 2a
Ê b2 (1 + tan2q) ˆ S
ËÁ b2 + a 2 tan 2q ˜¯
= 2a
2ab2 Q
a2sin2q + b2cos2q
= The equation of the chord joining the points P(a) and
Hence the result. P(b) is
(vi) If P(a) and P(b) are the extremities of a focal chord, x cos ÊÁË a+ b ¯˜ˆ + y sin ÁËÊ a+ b ˜¯ˆ = cos ËÊÁ a- b ˆ¯˜
a 2 b 2 2
then
which is passing through (ae, 0),
ÊÁË a ¯˜ˆ Ê bˆ e -1
tan 2 tan ËÁ 2 ¯˜ = e +1 . We have,
Proof e cos Ê a+ bˆ = cos Ê a — bˆ
ËÁ 2 ¯˜ ËÁ 2 ˜¯
P(a)
fi e ¥ 2 sin ÊËÁ a + b ˜¯ˆ cos ÊÁË a + b ¯ˆ˜
2 2
= 2 sin ËÊÁ a + b ¯˜ˆ cos Ê a - b ˆ
2 ÁË 2 ˜¯
Q(b)
Let the equation of the ellipse be fi e ¥ sin (a + b) = sin a + sin b
fi e = sin a + sin b
x2 + y2 = 1.
a2 b2 sin (a + b)
Hence, the result.
Equation of the chord PQ is
x cos ÁËÊ a + b ˜¯ˆ + y sin ÊËÁ a + b ¯˜ˆ = cos ËÁÊ a - b ˆ˜¯ 7. POSITION OF A POINT WITH RESPECT TO AN ELLIPSE
a 2 b 2 2
which is passing through the focus (ae, 0), then The point (x1, y1) lies outside, on or inside the ellipse
e cos ÁËÊ a + b ¯˜ˆ = cos ÊËÁ a - b ¯˜ˆ x2 + y2 =1 according as
2 2 a2 b2
5.6 Coordinate Geometry Booster
x12 + y12 -1> ,=,< 0 10. VARIOUS FORMS OF TANGENTS
a2 b2
(i) Point form
8. INTERSECTION OF A LINE AND AN ELLIPSE The equation of the tangent to the ellipse x2 + y2 =1
at (x1, y1) is a2 b2
The line y = mx + c intersects the ellipse x2 + y2 = 1 into two xx1 + yy1 =1
a2 b2 a2 b2
(i) real and distinct points if x2 < a2m2 + b2.
(ii) coincident points if c2 = a2m2 + b2. (ii) Parametric form x2 y2
(iii) imaginary points if c2 > a2m2 + b2. a2 b2
The equation of the tangent to the ellipse + =1
at (a cos q, b sin q) is
Also
(iv) The line y = mx + c will be a tangent to the given ellipse x cos q + y sin q = 1
if ab
c2 = a2m2 + b2.
(iii) Slope form
(v) The co-ordinates of the point of contact is x2 y2
a2 b2
Ê a 2m b2 ˆ The equation of the tangent to the ellipse + =1
ÁË c c ¯˜ in terms of the slope m is
± , ± which is also known as the m-point on
the ellipse. y = mx + a2m2 + b2
(vi) The equation of any tangent to the ellipse can be con- (iv) The co-ordinates of the points of contact are
sidered as y = mx + a2m2 + b2 .
Ê a2m , ± b2 ˆ
(vii) The line lx + my + n = 0 will be a tangent to the ellipse Á a2m2 + b2 a2m2 b2 ˜ .
Ë ¯
+
x2 + y2 = 1, if a2l2 + b2m2 = n2 (v) The point of intersection of the tangents at P(q) and
a2 b2 Q(j) is
9. THE LENGTH OF THE CHORD INTERCEPTED Ê a cos ÊËÁ q + j¯ˆ˜ b sin Ê q + jˆ ˆ
BY THE ELLIPSE ON THE LINE Y = mx + C. Á 2 ÁË 2 ˜¯ ˜
Á , ˜
ËÁÁ Ê q — jˆ Ê q — jˆ ˜¯˜
Let the equation of the ellipse be cos ËÁ 2 ¯˜ cos ÁË 2 ˜¯
x2 + y2 =1
a2 b2
11. DIRECTOR CIRCLE
x2 (mx + c)2
fi a2 + b2 =1 The locus of the point of intersec- P(h, k)
tion of two perpendicular tangents
fi (a2m2 + b2)x2 + 2a2mcx + a2(c2 – b2) = 0 to an ellipse is known as the direc-
tor circle.
Let its roots are x1, x2.
2a2mc a2 (c2 - b2 ) The equation of the director cir-
a2m2 + b2 a2 m2 + b2
Then x1 + x2 = - and x1 ◊ x2 = . x2 y2 is x2 +
a2 b2
cle to an ellipse + =1
y2 = a2 + b2.
2ab a2m2 + b2 - c2
Thus, x1 - x2 = .
a2m2 + b2
12. PAIR OF TANGENTS
Hence the length of the chord Equation of a pair of tangents from a point (x1, y1) to an
= (x1 - x2 )2 + ( y1 - y2 )2 ellipse x2 + y2 = 1 is
a2 b2
= (x1 - x2 ) 1 + m2
= 2ab ¥ 1 + m2 a2m2 + b2 - c2 Ê x2 + y2 - ˆ Ê x12 + y12 ˆ = Ê xx1 + yy1 - 1ˆ¯˜ 2
a2m2 + b2 ÁË a2 b2 1¯˜ ÁË a2 b2 - 1¯˜ ËÁ a2 b2
Ellipse 5.7
P Let Q(a, b) be any point in the xy-plane.
Equation (i) passes through Q(a, b), so we have
aa sec q – bb cosec q = a2 – b2
QR Ê 1 + tan 2 Ê qˆ ˆ Ê 1 + tan 2 Ê qˆ ˆ
Á tan 2 ÁË 2 ¯˜ ˜ Á ËÁ 2¯˜ ˜
fi aaÁ Ê qˆ ˜ - bbÁ qˆ ˜ = a2 - b2
ËÁ 2 ¯˜ ˜˜¯ Ê 2 ˜¯ ˜˜¯
ÁËÁ 1 - ÁÁË 2 tan ÁË
13. VARIOUS FORMS OF NORMALS fi 2aaÊÁË1 + tan 2 Ê qˆˆ tan Ê qˆ - b b ÊÁË1 - tan 4 Ê qˆˆ
ÁË 2˜¯ ˜¯ ÁË 2 ˜¯ ÁË 2˜¯ ˜¯
P T
= 2(a2 - b2 ) tan Ê qˆ ËÁÊ1 - tan 2 Ê qˆ ˆ
ËÁ 2 ˜¯ ÁË 2 ˜¯ ¯˜
fi bbt4 + 2(a2 – b2 + aa)t3 –2(a2 – b2 – aa)t – bb = 0
Ê qˆ …(ii)
ÁË 2 ˜¯
where t = tan
N
Here, PT be a tangent and PN be a normal. The above equation will give four values of t say, t1, t2, t3, t4.
Corresponding to these four values of t, we will get 4
The angle between the tangent and the normal is 90°.
points, say A, B, C, D on the ellipse, the normals to which
(i) Point form x2 y2 pass through the point Q(a, b).
The equation of the normal to the ellipse a2 b2
+ =1 Hence, in general four normals can be drawn from any
at the point (x1, y1) is point to an ellipse.
a2x - b2 y = a2 - b2 . (i) Co-normal points
x1 y1 Let P, Q, R, S are four points on the ellipse. If the nor-
mals at these points meet at a point, say M, then these
(ii) Parametric form x2 y2 four points are known as co-normal points.
a2 b2
The equation of the normal to an ellipse + =1 at
the point (a cos q, b sin q) is
RQ
ax sec q – by cosec q = a2 – b2.
(iii) Slope form M
The equation of the normal in terms of slope is
y = m x m(a2 - b2) P
a2 + b2m2 S
x2 y2
(iv) The line y = mx + c is a normal to an ellipse a2 + b2 =1 (ii) If a, b, g, d be the eccentric angles of the four points
if on the ellipse such that the normals at these points are
concurrent, then a + b + g + d = (2n + 1) p, n Π1.
c2 = m2(a2 - b2)2 . As we know that if four normals are concurrent at a
(a2 + b2m2) point, say, M(a, b), then
(v) The straight line lx + my + n = 0 is a normal to an bbt2 + 2(a2 – b2 + aa)t3 –2(a2 – b2 – aa)t = bb = 0,
ellipse x2 + y2 = 1 if where t = tan Ê qˆ
a2 b2 ÁË 2 ˜¯
a2 + b2 = (a2 - b2)2 Now, tan ÊËÁ a + b + g + 2d ¯ˆ˜ = S1 - S3
l2 m2 n2 2 2 2 1 - S2 + S4
14. NUMBER OF NORMALS ARE DRAWN TO AN - 2[aa + (a2 - b2 )] + 2[ab - (a2 - b2 )]
ELLIPSE FROM A POINT TO ITS PLANE = bb bb
x2 y2 1- 0 -1
a2 b2
The equation of the normal to the ellipse + = 1 at P(q) is =
ax sec q – by cosec q = a2 – b2 …(i) fi cot Ê a + b + g + dˆ = 0
ËÁ 2 2 2 2 ¯˜
5.8 Coordinate Geometry Booster
Ê a b g dˆ Ê 2n + 1ˆ Properties related to pole and polar
ÁË 2 2 2 2 ¯˜ ËÁ 2 ˜¯
fi cot + + + = 0 = cot p (i) The polar of the focus is the directrix.
Êa b g dˆ Ê 2n + 1ˆ (ii) Any tangent is the polar of the point of contact.
ÁË 2 2 2 2 ¯˜ ÁË 2 ˜¯
fi + + + = p (iii) The pole of a line lx + my + n = 0 with respect to the
fi a + b + g + d = (2n + 1)p. ellipse x2 + y2 =1 is Ê - a2l , - b2m ˆ .
a2 b2 ËÁ n n ¯˜
Hence, the result. (iv) The pole of a given line is the same as the point of in-
15. CHORD OF CONTACT tersection of tangents at its extremities.
P (v) If the polar of P(x1, y1) passes through Q(x2, y2), the
R polar of Q(x2, y2) goes through P(x1, y1) and such points
are said to be conjugate points.
Q
(vi) If the pole of a line lx + my + n = 0 lies on the another
line l¢x + m¢y + n¢ = 0, the pole of the second line will lie
on the first and such lines are said to be conjugate lines.
Chord of contact 18. DIAMETER
The equation of the chord of contact of tangents drawn from The locus of the mid-points of a system of parallel chords of
an ellipse is called a diameter and the point where the diam-
a point (x1, y1) to an ellipse eter intersects the ellipse is called the vertex of the diameter.
x2 + y2 = 1 is xx1 + yy1 = 1 B
a2 b2 a2 b2
16. CHORD BISECTED AT A GIVEN POINT M(h, k)
The equation of the chord bi- B A¢ A
M(h, k)
sected at a point (x1, y1) to an
ellipse x2 + y2 =1 is A y = mx + c
a2 b2 B¢
xx1 + yy1 -1= x12 + y12 -1 Let (h, k) be the mid-point of the chord, then
a2 b2 a2 b2
y = mx + c of the ellipse x2 + y2 =1.
a2 b2
i.e. T = S1
Then T = S1
fi
17. POLE AND POLAR xh + yk = h2 + k2
a2 b2 a2 b2
The equation of the polar of an ellipse
b2h
x2 + y2 =1 fi k = - a2m
a2 b2
Hence, the locus of the mid-point is y = - b2x .
from a point (x1, y1) is a2m
xx1 + yy1 =1, 19. CONJUGATE DIAMETERS
a2 b2
where SP
(x1, y1) is the pole of polar. C
Q
A B QR
P
Polar Two diameters are said to be conjugate when each bisects all
C
chords parallel to the other.
Q¢ D If y = m1x and y = m2x be two conjugate diameters of an
ellipse, then m1m2 = - b2 .
a2
Ellipse 5.9
Let PQ and RS be two conjugate diameters. Proof
Then the co-ordinates of the four extremities of two con-
jugate diameters are D P
C
P(a cos j, b sin j),
Let CP and CD be two conjugate semi-diameters of an
Q(–a cos j, – b sin j),
ellipse x2 + y2 =1 and the eccentric angle of P is f.
S(– a sin j, b cos j), a2 b2
Thus the eccentric angle of D is p + j .
and R(a sin j, –b cos j) 2
Properties of conjugate diameters Therefore the co-ordinates of P and D are (a cos j, b
(i) Prove that the eccentric angles of the ends of a pair
of conjugate diameters of an ellipse differ by a right
angle.
Proof
R(f¢) P(f) sin j) and
Ê a cos Ê p + j¯ˆ˜ , b sin Ê p + jˆ˜¯ ˆ
ËÁ ÁË 2 ËÁ 2 ˜¯
C i.e. (–a sin j, b cos j).
Thus CP2 + CD2
QS
= (a2 cos2 j + b2 sin2 j) + (a2 sin2 j + b2 cos2 j)
Let PCQ and RCS be two conjugate diameters of an = a2 + b2
ellipse x2 + y2 = 1. Hence, the result
a2 b2
(iii) Prove that the product of the focal distances of a point
Then the co-ordinates are P(a cos j, b sin j) and
on an ellipse is equal to the square of the semi-diameter,
R(a cos j¢, a sin j¢).
which is conjugate to the diameter through the point.
Proof
DP
Now, m1 = Slope of CP = b tan j and
a
b S(–æ, 0) C S(æ, 0)
a
m2 = Slope of CR = tan j¢
Since the diameters PCQ and RCS are conjugate diam- Let CP and CD be the conjugate diameters of the
ellipse.
eters, then Let P = (a cos j, b sin j), then the co-ordinates of D is
(–a sin j, b cos j).
m1 ◊ m2 = - b2 Thus,
a2
SP ◊ S¢P = (a – ae cos j) ◊ (a + ae cos j)
fi b2 tan j tan j¢ = - b2 = a2 – a2e2 cos2 j
a2 a2 = a2 – (a2 – b2) cos2 j
= a2 sin2 j + b2 cos2 j
fi tan j tan j¢ = –1 = CD2
fi tan j = - cot j¢ = tan ËÊÁ p + j¢ˆ¯˜ Hence, the result.
2
(iv) Prove that the tangent at the extremities of a pair of
fi j = p + j¢
2 conjugate diameters form a parallelogram whose area
fi j -j¢ = p is constant and equal to the product of the axes.
2
Proof
Hence, the result.
(ii) Prove that the sum of the squares of any two conjugate Y
semi-diameters of an ellipse is constant and equal to RP
the sum of the squares of the semi-axes of the ellipse,
i.e. X¢ X
C
CP2 + CD2 = a2 + b2.
QB
Y¢
5.10 Coordinate Geometry Booster
Let PCQ and RCS be two conjugate diameters of the (v) Equi-conjugate diameters
ellipse x2 + y2 = 1. B P
a2 b2 R
Then the co-ordinates of P, Q, R, and S are A¢ A
P(a cos j, b sin j), Q(–a cos j, – b sin j), C
R(–a sin j, b cos j) and S(a sin j, –b cos j) respec-
tively. B¢
Equations of tangents at P, R, Q and S are Two conjugate diameters are said to be equi-conjugate
diameters if their lengths are equal.
x cos j + y sin j = 1, i.e. CP = CR = a2 + b2
ab
2
- x sin j + y cos j = 1,
ab
- x cos j - y sin j = 1 20. REFLECTION PROPERTY OF AN ELLIPSE
ab
If an incoming light ray passes through one focus (S) strikes
and x sin j - y cos j = 1 the concave side of the ellipse, it will get reflected towards
ab other focus.
Thus, the tangents at P and Q are parallel. Y¢
Also the tangents at R and S are are parallel. P
Hence, the tangents at P, R, Q, S form a parallelogram.
Area of the parallelogram = MNM¢N¢
= 4(the area of the parallelogram CPMR)
= 4 ¥ a2cos2j + b2sin2j ¥ ab X¢ S¢ C S X
= 4ab a2cos2j + b2sin2j
= constant
Hence, the result. Y¢
EXERCISES
LEVEL I 5. Find the equation of the ellipse whose axes are co-ordi-
(Problems based on Fundamentals) nate axes and foci are (±2, 0) and the eccentricity is 1/2.
ABC OF AN ELLIPSE 6. If the distance between the foci of an ellipse is equal
to the length of its latus rectum, the eccentricity is
1. Find the centre, vertices, co-vertices, lengths of major
and minor axes, eccentricity, lengths of latus rectum, 5 -1.
equation of directrices and the end-points of a latus 2
recta. 7. Find the eccentricity of the ellipse x2 + y2 = 1 whose
(i) 9x2 + 16y2 = 144 a2 b2
(ii) 2x2 + 3y2 – 4x – 12y + 8 = 0
latus rectum is half of its major axis.
2. Find the sum of the focal distances of any point on the
ellipse 16x2 + 25y2 = 400. 8. Find the eccentric angle of a point on the ellipse
x2 + y2 = 1 whose distance from the centre of the
3. If the equation x2 + y2 = 1 represents of an 62
10 - a a — 4 ellipse is 5 .
ellipse such that the length of the interval, where a lies, 9. Find the area of the greatest rectangle that can be in-
scribed in the ellipse x2 + y2 = 1 .
is m, find m. 16 9
4. If (5, 12) and (24, 7) are the foci of an ellipse passing
10. Find the locus of a point whose co-ordinates are given
through the origin, find the eccentricity of the ellipse.
by x = 3 + 4 cos q cy = 2 + 3 sin q.
11. If PSQ is a focal chord of an ellipse 16x2 + 25y2 = 400
such that SP = 8, find the length of SQ.
Ellipse 5.11
12. Find the area bounded by the curve x2 + y2 £ 1 and 28. If p be the length of perpendicular from the focus S of
the line x + y ≥ 1. 16 9 the ellipse x2 + y2 =1 on the tangent at P, prove that
a2 b2
43
b2 2a
p2 = SP -1.
POSITION OF A POINT W.R.T. AN ELLIPSE 29. If p be the perpendicular from the centre of an
13. Find the location of the point (2, 3) with respect to the ellipse upon the tangent at any point P on it and
ellipse x2 + y2 = 1 .
43 r be the distance of P from the centre, prove that
14. If (l, –l) be an interior point of an ellipse 4x2 + 5y2 = 1 a2b2 = a2 + b2 - r2 .
such that the length of the interval, where l lies, is m, p2
where m ΠQ+, find the value of (3m Р2)2013 + 2013.
30. Prove that the locus of the mid-points of the portion
TANGENT AND TANGENCY of the tangents to the ellipse x2 + y2 =1 intercepted
a2 b2
a2 b2
15. Find the number of tangents drawn from a point (2, 3) between the axes is x2 + y2 =4 .
to an ellipse 4x2 + 3y2 = 12.
16. Find the equations of the tangents drawn from the point 31. Prove that the portion of the tangent to the ellipse inter-
(2, 3) to the ellipse 9x2 + 16y2 = 144.
cepted between the curve and the directrix subtends a
17. If the line 3x + 4y = 5 touches the ellipse 9x2 + 16y2 =
144, find the points of contact. right angle at the corresponding focus.
32. Prove that in an ellipse, the perpendicular from a focus
upon any tangent and the line joining the centre of the
18. For what value of l does the line y = x + l touches the ellipse to the point of contact meet on the correspond-
ellipse 9x2 + 16y2 = 144? ing directrix.
19. Find the equations of the tangents to the ellipse 33. Prove that the co-ordinates of those points on the
x2 + y2 = 1 having slope 2. ellipse x2 + y2 = 1 tangents at which make equal an-
34 a2 b2 is
20. Prove that the locus of the feet of perpendicular drawn Ê a2 , ± b2 ˆ
gles with the axes Á± a2 + b2 ˜ .
from the centre upon any tangent to the given ellipse is Ë a2 + b2 ¯
r2 = a2 cos2 q + b2 sin2 q.
34. Find the locus of the point of intersection of two per-
21. A circle of radius r is concentric with the ellipse pendicular tangents to the ellipse x2 + y2 = 1 .
16 9
x2 + y2 = 1. Prove that the common tangent is inclined
a2 b2
Ê r2 - b2 ˆ 35. Tangents are drawn from any point P on the parabola
to the major axis at an angle of tan-1 ÁË a2 - r2 ˜¯ .
(y – 2)2 = 4(x – 1) to the ellipse x2 + y2 = 1 , which are
22. Prove that the tangents at the extremities of latus rectum 41
of an ellipse intersect on the corresponding directrix.
mutually perpendicular to each other. Find the locus of
23. Prove that the locus of the mid-points of the portion of the point P.
the tangents to the given ellipse intercepted between
the axes is 4r2 = a2 sin2 q + b2 cos2 q. 36. Find the equations of the pair of tangents to the ellipse
2x2 + 3y2 = 1 from the point (1, 1).
24. A tangent to the ellipse x2 + y2 =1 meets the ellipse
x2 + y2 = a +b in the a2 b2 and Q, prove that the 37. If the tangents are drawn from a point (1, 2) to the el-
lipse 3x2 + 2y2 = 5, find the angle between the tangents.
ab points P NORMAL AND NORMALCY
tangents at P and Q are at right angles. 38. Find the equation of the normal to the ellipse 4x2 + 9y2
25. Prove that the product of the perpendiculars drawn = 20 at ÊÁË1, 4ˆ .
from the foci upon any tangent to an ellipse is constant, 3¯˜
i.e. b2.
39. Find the equation of the normal to the ellipse 5x2 + 3y2
26. If an ellipse slides between two perpendicular straight = 137 at the point whose ordinate is 2.
lines, prove that the locus of its centre is a circle.
40. Find the equation of the normal to the ellipse
27. Prove that the locus of the feet of the perpendiculars x2 + y2 = 1 at the negative end of the latus rectum.
from foci upon any tangent to an ellipse is an auxiliary 94
circle.
5.12 Coordinate Geometry Booster
41. If the normal at the point P(q) to the ellipse 5x2 + 14y2 CHORD OF CONTACT/CHORD BISECTED AT A POINT
= 70 intersects it again at the point Q(2q), prove that 51. Prove that the locus of the point, the chord of contact
3 cos q + 2 = 0. x2 y2
a2 b2
42. The normal at an end of a latus rectum of an ellipse of tangents from which to the ellipse + =1
x2 + y2 =1 passes through one extremity of the minor subtends a right angle at the centre of the ellipse is
a2 b2
axis, then prove that e4 + e2 – 1 = 0. x2 + y2 = 1 + 1 .
a4 b4 a2 b2
43. Prove that the tangent and the normal at any point of
52. Prove that the locus of the point, from which the chord
an ellipse bisect the the external and internal angles be-
of contact of tangents are to be drawn to the ellipse
tween the focal distances of the point.
44. The normal at any point P on the ellipse x2 + y2 =1 touches the circle x2 + y2 = c2 is x2 + y2 = 1 .
a2 b2 a4 b4 c2
meets the major and minor axes at G and G¢, respec- 53. The perpendicular tangents are drawn to the ellipse
tively and CF is perpendicular upon the normal from x2 y2
a2 b2
the centre C of the ellipse, show that PF ◊ PG = b2 and + =1. Prove that the locus of the mid-point of
the Ê x2
PF × PG¢ = a2. Ê x2 y2 ˆ 2 ËÁ a2 + y2 ˆ
The normal at a point P(q) on the ellipse ÁË a2 b2 ¯˜ + b2 ˜¯
45. x2 + y2 =1 chord of contact is + = .
a2 b2
54. Tangents are drawn from any point on the circle x2 + y2
meets the axes of x and y at M and N, respectively,
x2 y2
show that x2 + y2 = (a + b)2. x2 y2 = c2 to the ellipse a2 + b2 = 1. Prove that the locus of
a2 b2
46. An ordinate PN of an ellipse + =1 meets the the mid-points of the chord of contact is
auxiliary circle in Q. Show that the locus of the point Ê x2 + y2 ˆ2 = Ê x2 + y2 ˆ
ËÁ c2 b2 ˜¯ ÁË c2 ˜¯
of intersection of the normals at P and Q is the circle x2
+ y2 = (a + b)2. 55. Tangents PA and PB are drawn from a point P to the
47. Prove that the tangent of the angle between CP and the x2 y2
a2 b2
normal at P(q) is Ê a2 - b2 ˆ ¥ sin 2q and its greatest ellipse + = 1. The area of the triangle formed
ÁË 2ab ¯˜
by the chord of contact AB and axes of co-ordinates are
Ê a2 - b2 ˆ
value is ÁË 2ab ˜¯ . constant. Prove that the locus of P is a hyperbola.
56. Prove that the locus of the mid-points of the normal
48. Prove that in an ellipse, the distance between the centre chords of the ellipse x2 + y2 = 1 is
a2 b2
and any normal does not exceed the difference between
the semi-axes of the curve. Ê a6 b6 ˆ Ê x2 y2 ˆ 2
ÁË x2 y2 ¯˜ ËÁ a2 b2 ¯˜
49. The tangent and the normal at any point P of an el- + + = (a2 - b2)2
lipse x2 + y2 =1 cut its major axis in points Q and R, 57. Prove that the locus of the mid-points of the chords
a2 b2
x2 y2
respectively. If QR = a, show that the eccentric angle of of the ellipse a2 + b2 =1 which touch the auxiliary
the point P is satisfying the equation circle x2 + y2 = a2 is
e2 cos2 j + cos j – 1 = 0
x2 y2 ˆ Ê x2 y2 ˆ 2
50. If the normals at P(x1, y1), Q(x2, y2) and R(x3, y3) on a 2 Ê a4 + b4 ¯˜ = ËÁ a2 + b2 ¯˜
ËÁ
the ellipse x2 y2 are concurrent, show that
a2 + b2 =1 58. Prove that the locus of the mid-points of the chords of
x1 y1 x1y1 the ellipse x2 + y2 =1 which subtends a right angle at
x2 y2 x2 y2 = 0 a2 b2
x3 y3 x3 y3 the centre of the ellipse is
and if points P(a), Q(b) and R(g), prove that Ê x2 + y2 ˆ = Ê x2 + y2 ˆ 2 Ê 1 + 1ˆ
ÁË a4 b4 ¯˜ ÁË a2 b2 ¯˜ ËÁ a2 b2 ¯˜
sec a cosec a 1 59 The eccentric angles of two points P and Q on the
sec b cosec b 1 = 0 ellipse differ by p/2. Prove that the locus of the mid-
sec g cosec g 1 point of PQ is x2 + y2 = 2.
a2 b2
Ellipse 5.13
60. Prove that the locus of the mid-point of the chord of 72. Find the co-ordinates of the four extremities of two
contact of the perpendicular tangents to the ellipse x2 + y2 =1.
a2 b2
x2 y2 Ê x2 y2 ˆ 2 Ê x2 + y2 ˆ conjugate diameters of an ellipse
a2 b2 ËÁ a2 b2 ˜¯ ÁË a2 + b2 ¯˜
+ =1 is + = . 73. Prove that the sum of the squares of any two conjugate
61. Prove that the locus of the mid-points of the focal semi-diameters of an ellipse is constant and equal to
chords of the ellipse Ê x2 + y2 ˆ =1 is the sum of the squares of the semi-axes of the ellipse,
ÁË a2 b2 ˜¯ i.e. CP2 + CD2 = a2 + b2.
b2x2 + a2y2 = ab2xe 74. Prove that the product of the focal distances of a point
62. Prove that the locus of the mid-points of the chords of on an ellipse is equal to the square of the semi-diam-
the ellipse Ê x2 + y2 ˆ =1 which are tangents to the eter, which is conjugate to the diameter through the
ÁË a2 b2 ¯˜ point. i.e. SP ◊ S¢P = CD2 .
75. Prove that the locus of the poles of the line joining the
ellipse x2 + y2 =1 is eccentricities of two conjugate diameters is the ellipse
a2 b2
x2 + y2 = 2.
2 a2 b2
Ê x2 y2 ˆ Ê x2a 2 y2b2 ˆ
ÁË a2 + b2 ¯˜ = ËÁ a4 + b4 ¯˜ 76. If P and D be the ends of the conjugate diameters of an
ellipse, find the locus of the mid-point of PD.
POLE AND POLAR 77. For the ellipse x2 + y2 =1, find the equation of the
a2 b2
63. Prove that the polar of the focus of an ellipse is the
diameter conjugate to ax – by = 0.
directrix.
78. Iadfniadtmhecext2p2eor+isndotyf22otfh=ien1fteobrresmeacettrti,hopenroeovxfetrthethmeaeittlileacips22soe+fstdbhax22e22=c+o1nb.yju22g=at1e
64. Find the pole of a given line lx + my + n = 0 with re-
spect to an ellipse x2 + y2 =1.
a2 b2
65. Prove that the pole of a given line is the same as the
point of intersection of tangents at its extremities.
66. Find the pole of the straight line x + 4y = 4 with respect 79. If CP and CD are the conjugate diameters of the ellipse
to the ellipse x2 + 4y2 = 4.
x2 y2
67. Find the locus of the poles of the tangents to a2 + b2 =1, prove that the locus of the orthocentre of
x2 + y2 =1 with respect to the concentric ellipse c2x2 the DPCD is 2(b2y2 + a2x2)2 = (a2 – b2)2 (b2y2 – a2x2)2 .
a2 b2
80. Prove that the tangents at the extremities of a pair of
+ d2y2 = 1.
conjugate diameters form a parallelogram whose area
68. The perpendicular from the centre of an ellipse
is constant and equal to the product of the axes.
x2 y2
a2 + b2 =1 on the polar of a point with respect to the 81. Show that the tangents at the ends of conjugate diam-
ellipse is constant and equal to c. Prove that the locus eters of the ellipse x2 + y2 = 1 intersect on the ellipse
a2 b2
x2 y2 1 x2 y2
of the point is the ellipse a4 + b4 = c2 . a2 + b2 = 2.
69. Show that the equation of the locus of the poles of nor- 82. Find the eccentricity of the ellipse if y = x and 2x + 3y =
0 are the equations of a pair of its conjugate diameters.
mal chords of the ellipse x2 + y2 = 1 is
a2 b2
REFLECTION PROPERTY OF AN ELLIPSE
(a2 – b2)2 x2y2 = a6y2 + b6x2
70. If the polar with respect to y2 = 4x touches the ellipse 83. A ray is emanating from the point (–3, 0) is incident on
the ellipse 16x2 + 25y2 = 400 at the point P with ordi-
x2 + y2 = 1, find the locus of its pole. nate 4. Find the equation of the reflected ray after first
a2 b2 reflection.
CONJUGATE DIAMETERS 84. A ray is coming along the line x – y + 2 = 0 on the
ellipse 3x2 + 4y2 = 12. After striking the elliptic mirror,
71. Prove that the equation of the diameter of the ellipse it is then reflected. Find the equation of the line con-
taining the reflected ray.
x2 y2 b2x
a2 + b2 = 1 is y = - a2m .
5.14 Coordinate Geometry Booster
LEVEL II 12. The equation of common tangent between the ellipses
(Mixed Problems)
x2 + y2 = 1 and x2 + y2 = 1 is
94 21
1. Tangents are drawn from a point on the circle x2 + y2
(a) x = 3 (b) y = 2
= 50 to the ellipse x2 + y2 = 1 , the tangents are at the (c) x = 1 (d) not defined
angle is 30 20 13. If the equation of the tangent to the ellipse x2 + y2 =1
is x + y = 2 , its eccentric angle is a2 b2
(a) p (b) p (c) p (d) p
4 3 2 8 ab
2. The centre of the ellipse 4x2 + 9y2 – 8x – 36y + 4 = 0 is (a) 45° (b) 60° (c) 90° (d) 22.5°
14. The equations of the tangents to the ellipse 3x2 + y2 = 3
(a) (2, 4) (b) (3, 2) (c) (1, 2) (d) (0, 1)
3. The co-ordinates of the foci of the ellipse 4x2 + 9y2 – 8x making equal intercepts on the axes are
– 36y + 4 = 0 is (a) y = ±x ±2 (b) y = ±x ±4
(a) (1 ± 5, 2) (b) (2 ± 5, 2) (c) y = ±x ±5 (d) y = ±x ±7
(c) (± 5, 2) (d) (1 ± 5, 3) 15. The number of real tangents can be drawn from (3, 5)
to the ellipse 3x2 + 5y2 = 15 is
4. The equation x2 + y2 = 1 represents an ellipse
if 10 - a 4 - a (a) 4 (b) 2 (c) 1 (d) 0
16. The number of normals that can be drawn from a point
(a) a < 4 (b) a > 4 to a given ellipse is
(c) 4 < a < 10 (d) a > 10 (a) 4 (b) 2 (c) 1 (d) 0
5. Lep P be a variable point on the ellipse x2 + y2 = 1 17. The set of possible values of m for which a line with the
25 16
slope m is a common tangent to the ellipse x2 + y2 =1
and the parabola y2 = 4ax is b2 c2
with foci at F and F¢. If A be the area of the DPFF¢, the
maximum value of A is (a) (3, 5) (b) (2, 3) (c) (1, 3) (d) (0, 1)
(a) 12 s.u. (b) 24 s.u. 18. The angle between the normals of the ellipse 4x2 + y2 =
(c) 36 s.u. (d) 48 s.u.
6. The eccentricity of the ellipse (10x – 5)2 + (10y – 5)2 = 5, at the intersection of 2x + y = 3 and ellipse is
(3x + 4y – 1)2 is
(a) tan–1(3/5) (b) tan–1(3/4)
(a) 1 1 (c) 1 (d) 1 (c) tan–1(4/3) (d) tan–1(4/5)
2 (b) 2 4
19. If the latus rectum of an ellipse x2 tan2 j + y2 sec2 j =
3
1 is 1/2, then f is
7. If the line y = x + l touches the ellipse 9x2 + 16y2 = 144,
the value of l is (a) p/2 (b) p/6 (c) p/3 (d) 5p/12
(a) ±5 (b) ±4 (c) ±7 (d) ±3 20. If pair of tangents are drawn to the ellipse x2 + y2 = 1
16 9
8. The equation of the tangents to the ellipse 3x2 + 4y2 =
12 which are perpendicular to the line y + 2x = 4 is/are from a point P, so that the tangents are at right angles to
each other, the possible co-ordinates of the point P are
(a) x – 2y + 4 = 0 (b) x – 2y + 7 = 0 (a) (3 2, 7) (b) (5, 0)
(c) x – 2y – 4 = 0 (d) x – 2y – 7 = 0 (c) (3, 4) (d) (2 5, 5)
9. The product of the perpendiculars from the foci of any 21. The eccentricity of the ellipse whose pair of a conju-
tangent to an ellipse b2x2 + a2y2 = a2b2 is
gate diameters are y = x and 3y = –2x is
(a) a2 (b) b2 (c) 2b2 (d) 2a2 (a) 1/ 3 (b) 2/3 (c) 1/3 (d) 1/5
10. The number of tangents to the ellipse x2 + y2 = 1 22. The minimum length of intercept of any tangent of an
from the point (4, 3) is 16 9 ellipse x2 + y2 =1 between the axes is
a2 b2
(a) 0 (b) 1 (c) 2 (d) 3
11. If the normal at an end of a latus rectum of an ellipse (a) 2a (b) 2b (c) a + b (d) a – b
x2 + y2 =1 passes through one extremity of the minor 23. The eccentric angle of a point on the ellipse x2 + y2 = 1
a2 b2 whose distance from the centre is 2, is 6 2
axis, then e2 is
(a) p/4 (b) p/6 (c) p/2 (d) p/6
(a) 3 - 1 (b) 3 + 1 24. If l2 x2 - 6 + l2 y2 + 5 = 1 represents an ellipse,
2 2 -l - 6l
(c) 5 + 1 (d) 5 - 1 then l lies in
2 2
Ellipse 5.15
(a) (– , –2) (b) (1, ) 34. An ellipse has OB as semi-minor axis, F, F¢ are its foci
and the angle FBF¢ is a right angle. The eccentricity of
(c) (3, ) (d) (5, ) the ellipse is
25. The eccentricity of the ellipse ax2 + by2 + 2gx + 2fy + c
= 0, if its axis is parallel to x-axis, is (a) 1/2 (b) 1/ 2 (c) 1/3 (d) 1/4
(a) a + b (b) a - b 35. The sum of the focal distances from any point on the
4 2 ellipse 9x2 + 16y2 = 144 is
(c) b - 1 (d) 1 - a (a) 32 (b) 18 (c) 16 (d) 8
a b
36. If P = (x, y), F1 = (3, 0), F2 = (–3, 0) and 16x2 + 25y2 =
26. S and T are the foci of an ellipse and B is an end of the 400, then PF1 + PF2 is equal to
minor axis. If STB is an equilateral triangle, the eccen- (a) 8 (b) 6 (c) 10 (d) 12
tricity of the ellipse is 37. The number of values of c such that the straight line
(a) 1/4 (b) 1/3 (c) 1/2 (d) 2/3 y = 4x + c touches the ellipse x2 + y2 = 1 is
4
27. The centre of the ellipse
(x + y - 1)2 + (x - y)2 = 1 is (a) 0 (b) 1 (c) 2 (d) infinite
9 16
38. The area of the quadrilateral formed by the tangents at
(a) (0, 0) (b) (1, 1) (c) (2, 1) (d) (1, 2) the points of latus recta to the ellipse x2 + y2 = 1 is
95
28. Let P be a variable point on the ellipse x2 + y2 =1
a2 b2 (a) 27/4 s.u. (b) 9 s.u.
with foci F1 and F2. If A be the area of the DPF1F2, the (c) 27/2 s.u. (d) 27 s.u.
maximum value of A is 39. Tangents are drawn to the ellipse x2 + y2 = 1 at
27
(a) ea (b) ab (c) aeb (d) e
b e ab (3 3 cos q, sin q ) , where 0 < q < p/2. The value of
29. If the angle between the lines joining the foci of an q for which the sum of intercepts on the axes made by
ellipse to an extremity of a minor axis is 90°, the this tangent is minimum is
eccentricity of the ellipse is (a) p/3 (b) p/4 (c) p/8 (d) p/6
40. If tangents are drawn to the ellipse x2 + 2y2 = 2, the
(a) 1/8 (b) 1/4 (c) 1/ 2 (d) 1/ 3
30. On the ellipse 4x2 + 9y2 = 1, the points at which the locus of the mid-point of the intercept made by the tan-
tangents are parallel to the line 8x = 9y are
gents between the co-ordinate axes is
(a) Ê 2 , 1ˆ (b) Ê - 2 , 1ˆ (a) 1 + 1 =1 (b) 1 + 1 =1
ËÁ 5 5¯˜ ÁË 5 5¯˜ 2x2 4y2 4x2 2y2
(c) Ê - 2 , - 1ˆ (d) Ê 2 , - 1ˆ (c) x2 + y2 = 1 (d) x2 + y2 = 1
ËÁ 5 5˜¯ ÁË 5 5¯˜ 24 42
31. The equation of the largest circle with centre (1, 0) that 41. The minimum area of the triangle formed by the tan-
can be inscribed in the ellipse x2 + 4y2 = 36 is gent to the ellipse x2 + y2 =1 and the co-ordinate
(a) 3x2 + 3y2 – 6x – 8 = 0 axes is a2 b2
(b) 3x2 + 3y2 + 6x – 8 = 0
(c) 2x2 + 2y2 + 5x – 8 = 0 (a) ab s.u. (b) a2 + b2 s.u.
(d) 2x2 + 2y2 – 5x – 8 = 0 2
32. Let E be the ellipse x2 + y2 =1 and C be the circle (c) (a + b)2 s.u. (d) a2 + ab + b2 s.u.
a2 b2 2 2
x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1) 42. The point on the curve x2 + 2y2 = 6 whose distance
from the line x + y = 7 is minimum, is
respectively. Then (a) (1, 2) (b) (1, 3) (c) (2, 1) (d) (3, 1)
(a) Q lies inside C but outside E 43. The equation of the common tangent in 1st quadrant to
the circle x2 + y2 = 16 and the ellipse x2 + y2 = 1 is
(b) Q lies outside both C and E 25 4
(c) P lies inside both C and E
(d) P lies inside C but outside E
33. The radius of the circle passing through the foci of the
ellipse x2 + y2 = 1 and having its centre at (0, 3) is (a) 2x + y 3 = 4 7 (b) 2x - y 3 = 4 7
16 9
(a) 4 (b) 3 (c) 5 (d) 7 (c) 3x - y 3 = 4 7 (d) 3x + y 3 = 4 7
5.16 Coordinate Geometry Booster
44. Let P(x1, y1) and Q(x2, y2), y1 < 0, be the end-points of 50. The number of points on the ellipse x2 + y2 = 1 from
the latus rectum of the ellipse x2 + 4y2 = 4. The equa- 50 20
tion of the parabolas with latus rectum PQ are
(a) x2 + 2 3y = 3 + 3 (b) x2 - 2 3y = 3 + 3 which pair of perpendicular tangents are drawn to the
(c) x2 + 2 3y = 3 - 3 (d) x2 - 2 3y = 3 - 3
ellipse x2 + y2 = 1 is
45. The line passing through the extremity A of the major 16 9
axis and extremity B of the minor axis of the ellipse
x2 + 9y2 = 9 meets the auxiliary circle at the point M. (a) 0 (b) 2 (c) 1 (d) 4
The area of the triangle with vertices at A, M and the
origin is 51. The eccentricity of the ellipse (x - 3)2 + ( y - 4)2 = y2
is 9
(a) 31 (b) 29 (c) 21 (d) 27 (a) 3 (b) 1 (c) 1 (d) 1
10 10 10 10 2 3 32 3
46. The normal at a point P on the ellipse x2 + 4y2 = 16 52. For an ellipse x2 + y2 = 1 with vertices A and A¢,
meets the x-axis at Q. If M is the mid-point of the seg- 94
ment PQ, the locus of M intersects the latus rectum of
the given ellipse at the points tangents are drawn at the point P in the first quadrant
meets the y-axis in Q and the chord A¢P meets the
y-axis in M. If O be the origin, then OQ2 – MQ2 is
(a) Ê ± 3 5 , ± 2ˆ (b) Ê ± 3 5 , ± 19 ˆ (a) 9 (b) 13 (c) 4 (d) 5
ËÁ 2 7 ˜¯ ÁË 2 4 ˜¯
53. The line lx + my + n = 0 will cut the ellipse x2 + y2 =1
a2 b2
ÁËÊ ± 2 1 ˆ˜¯ Ê 3, ± 4 3 ˆ
(c) 3, ± 7 (d) ËÁ ± 2 ˜¯ in points whose eccentric angles differ by p/2, if
7
(a) a2l2 + b2n2 = 2m2 (b) a2m2 + b2l2 = 2n2
47. Tangents are drawn from the point P(3, 4) to the ellipse
x2 + y2 = 1 touching the ellipse at A and B. Then the (c) a2l2 + b2m2 = 2n2 (d) a2n2 + b2m2 = 2l2.
94
54. A circle has the same centre as an ellipse and passes
co-ordinates of A and B are
through the foci F1 and F2 of the ellipse such that two
(a) (3, 0) and (0, 2) curves intersect in 4 points. Let P be any of their points
of intersection. If the major axis of the ellipse is 17 and
(b) Ê - 8 , 2 161ˆ and Ê - 9, 8ˆ
ÁË 5 15 ¯˜ ËÁ 5 5˜¯ the area of the triangle F1F2 is 30, the distance between
the foci is
Ê 8 2 161ˆ (a) 11 (b) 12 (c) 13 (d) None
ÁË 5 15 ˜¯
(c) - , and (0, 2) 55. The point O is the centre of the ellipse with major axis
AB and minor axis CD and the point F is one focus of
(d) (3, 0) and Ê - 9 , 8ˆ the ellipse. If OF = 6 and the diameter of the inscribed
ÁË 5 5˜¯
circle of DOCF is 2, the product of (AB)(CD) is
48. The maximum area of an isosceles triangle inscribed in (a) 65 (b) 52 (c) 78 (d) none
an ellipse x2 + y2 =1 with its vertex at one end of the 56. A tangent having slope –4/3 to the ellipse x2 + y2 = 1
a2 b2 18 32
major axis is intersects the major and minor axes in points A and B,
(a) 3 3 ab (b) 3 2 ab respectively. If C is the centre of the ellipse, the area of
4 4
the triangle ABC is
(a) 12 s.u. (b) 24 s.u.
(c) 2 3 ab (d) 3 3 ab (c) 36 s.u. (d) 48 s.u.
5 5
57. The common tangent to the ellipse x2 + y2 =1
a2 + b2 b2
49. The point (a, b) on the ellipse 4x2 + 3y2 = 12, in the x2 y2
first quadrant, so that the area enclosed by the lines and a2 + a2 + b2 =1 is
y = x and y = b, x = a and the x-axis is maximum, is
(a) ay = bx + a4 - a2b2 + b4
(a) Ê 3 , 1˜ˆ¯ (b) ÁËÊ1, 3ˆ (c) Ê 2, 3ˆ (d) Ê 3, 3ˆ (b) by = ax - a4 + a2b2 + b4
ÁË 2 2˜¯ ËÁ 2¯˜ ÁË 2˜¯
Ellipse 5.17
(c) ay = bx - a4 + a2b2 + b4 (c) ( , 7) Ê 7 , 12ˆ
ÁË 5 ˜¯
(d) by = ax + a4 - a2b2 + b4
(d) no such value of a exists
58. The normal at a variable point P on an ellipse 66. If the ellipse x2 + y2 =1 is inscribed in a rectangle
a2 b2
x2 + y2 =1 of eccentricity e meets the axes of the el-
a2 b2 whose length : breadth is 2 : 1, the area of the rectangle
lipse in Q and R, the locus of the mid-point of QR is a is
conic with an eccentricity e¢ such that (a) 4 (a2 + b2 ) (b) 4 (a2 + b2 )
7 3
(a) e¢ is independent of e
(b) e¢ =1 (c) e¢ = e (d) e¢ = 1/e (c) 12 (a2 + b2 ) (d) 8 (a2 + b2 )
5 5
59. An ellipse is drawn with major and minor axes of
lengths 10 and 8, respectively. Using one focus as the 67. The length of the side of the square which can be
centre, a circle is drawn that is the tangent to the el- made by four perpendicular tangents to the ellipse
lipse, and no part of the circle being outside of the el- x2 + y2 = 1 is
7 11
lipse. The radius of the circle is
(a) 3 (b) 2 (c) 2 2 (d) 5 (a) 10 (b) 8 (c) 6 (d) 5
60. A common tangent to 9x2 + 16y2 = 144; y2 = x – 4 and 68. If the tangent to the ellipse x2 + y2 =1 makes angles
x2 = y2 – 12x + 32 = 0 is a2 b2
(a) y = 3 (b) x + 4 = 0 a and b with the major axis such that tan a + tan b = l,
(c) x = 4 (d) y + 3 = 0 the locus of their points of intersection is
61. The area of the rectangle formed by the perpendiculars (a) x2 + y2 = a2 (b) x2 + y2 = b2
from the centre of the standard ellipse to the tangent (c) x2 = 2llx = a2 (d) l(x2 – a2) = 2xy.
and the normal at its point whose eccentric angle is p/4
is 69. If a – b = c, the locus of the points of intersection of
tangents at
(a) Ê a 2 - b 2 ˆ ab Ê a2 - b2 ˆ 1 P(a cos a, b sin a) and Q(a cos b, b sin b)
ÁË a 2 + b 2 ˜¯ (b) ËÁ a2 + b2 ¯˜ ab
(c) Ê a2 + b2 ˆ 1 (d) Ê a2 + b2 ˆ ab to the ellipse x2 + y2 = 1 is
ÁË a2 - b2 ¯˜ ab ËÁ a2 - b2 ¯˜ a2 b2
(a) a circle (b) a straight line
62. An ellipse having foci at (3, 3) and (–4, 4) and passing (c) an ellipse (d) a parabola
through the origin has eccentricity equal to 70. If the eccentricity of the ellipse x2 + y2 =1 is
a2 + 1 a2 +
(a) 3/7 (b) 2/7 (c) 5/7 (d) 3/5 2
63. A bar of length 20 units moves with its ends on two 1 , the latus rectum of the ellipse is
fixed straight lines at right angles. A point P marked on 6
the bar at a distance of 8 units from one and describes a (a) 5 (b) 10 (c) 8 (d) 7
6 6 6 6
conic whose eccentricity is
(a) 5/9 (b) 2 (c) 4 (d) 3/5
3 9
64. If maximum distance of any point on the ellipse LEVEL III
x2 + 2y2 + 2xy = 1 from its centre be r, the value of r is (Problems for JEE Advanced)
(a) 3 + 3 (b) 2 + 2 x2 y2
a2 b2
(c) 2 (d) 2 - 2 1. Let P be a variable point on the ellipse + =1
3- 5
with foci F1 and F2. If A be the area of DPF1F2, find the
65. If the ellipse x2 7 + y2 = 1 is inscribed in a maximum value of A.
a2 - 13 - 5a
2. Let d be the perpendicular distance from the centre of
square of side length a 2 , then a is x2 y2
a2 b2
(a) 6 the ellipse + = 1 to the tangent drawn at a point
5
, 7) Ê 7 , 13ˆ P on the ellipse. If F1 and F2 be the two foci of the
(b) ( ËÁ 5 ¯˜ ellipse, prove that
5.18 Coordinate Geometry Booster
(PF1 - PF2 )2 = 4a2 Ê - b2 ˆ 16. Find the equation of the common tangent in first
ËÁ1 d2 ¯˜ quadrant to the circle x2 + y2 = 16 and the ellipse
3. Find the radius of the circle passing through the foci of x2 + y2 = 1. Also find the length of the intercept of
25 4
the ellipse x2 + y2 = 1 and having its centre at (0, 3).
16 9 the tangent between the co-ordinate axes.
17. P and Q are two points on the ellipse x2 + y2 = 1 such
4. If a tangent drawn at a point (t2, 2t) on the parabola
25 9
y2 = 4x is the same as the normal drawn at a point
that sum of their ordinates is 3. Find the locus of the
( 5 cos j, 2 sin j) on the ellipse 4x2 + 5y2 = 20. Find points of intersection of tangents at P and Q.
18. From any point P lying in the first quadrant on the
the values of t and j. ellipse x2 + y2 = 1 , PN is drawn perpendicular to the
5. An ellipse has OB as a minor axis. F and F¢ are its foci 25 16
and the angle FBF¢ is a right angle. Find the eccentric-
ity of the ellipse. major axis and produced at Q so that NQ equals to PS,
6. A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse where S is a focus. Prove that the locus of Q is 3x + 5y
x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and
+ 25 = 0.
Q of the ellipse x2 + 2y2 = 6 are at right angles. x2 y2
19. If a tangent of slope 2 of the ellipse a2 + b2 = 1 is the
7. Find the co-ordinates of the points P on the ellipse
x2 + y2 = 1, for which the area of the DPON is maxi- normal to the circle x2 + y2 + 4x + 1 = 0, find the maxi-
a2 b2
mum value of ab.
mum, where O denotes the origin and N, the foot of the 20. If the maximum distance of any point on the ellipse
x2 + 2xy + 2y2 = 1 from its centre be r, find r.
perpendicular from O to the tangent at P.
8. Find the angle between the pair of tangents drawn to 21. Prove that the locus of the mid-points of the chords of
the ellipse 3x2 + 2y2 = 5 from the point (1, 2).
the ellipse x2 + y2 =1 which are of constant length 2c
9. If the normal at P(q) to the ellipse x2 + y2 = 1 inter- a2 b2
14 5
is Ê b2x2 + a2 y2 ˆ = 1 Ê x2 - y2 ˆ .
sects it again at Q(2q), prove that cos q = — 2 . ËÁ a4 y2 + b4x2 ¯˜ c2 ËÁ1 - a2 b2 ˜¯
3
22. Chords of the ellipse touch the parabola ay2 = –2b2x.
10. If Ê 1 , 2 ˆ be the mid-point of the chord of the ellipse
ËÁ 5 5 ¯˜ Prove that the locus of their poles is the parabola
ay2 = 2b2x.
x2 + y2 = 1 , find its length.
25 16 23. An ellipse is rotated through a right angle in its own
plane about its centre, which is fixed. Find the locus of
11. Prove that, in an ellipse, the perpendicular from a focus the point of intersection of a tangent to the ellipse in its
original position with the tangent at the same point of
upon any tangent and the line joining the centre of the the curve in its new position.
ellipse to the point of contact meet on the correspond-
ing directrix. 24. The tangents drawn from a point P to the ellipse make
angles q1 and q2 with the major axis. Find the locus of
12. Find the area of the quadrilateral formed by the tan- P for which q1 + q2 = 2a.
gents at the end-points of latus rectum to the ellipse 25. Find the length of the sides of a square which can be
x2 + y2 = 1. x2 + y2 = 1 made by four perpendicular tangents to the ellipse
95 27
x2 + y2 = 1.
13. A tangent is drawn to the ellipse at 16 9
(3 3 cos q, sin q ) , where 0<q < p .
2
26. Find the locus of the point which divides the double
Find the value of q such that the sum of the intercepts
ordinates of the ellipse x2 + y2 = 1 in the ratio 1 : 2
on axes made by the tangent is minimum. 94
14. If tangents are drawn to the ellipse x2 + 2y2 = 2, find internally.
the locus of the mid-point of the intercept made by the 27. An ellipse has the points (1, –1)and (2, –1) as its foci
and x + y = 5 as one of its tangent. Find the point where
tangents between the co-ordinate axes. the line be a tangent to the ellipse from the origin.
15. Find the minimum area of the triangle formed by the 28. An ellipse is sliding along the co-ordinate axes. If the
foci of the ellipse are (1, 1) and (3, 3), find the area of
tangent to the ellipse x2 + y2 = 1 and the co-ordinate the director circle of the ellipse.
axes. a2 b2
Ellipse 5.19
29. What are the values of the parameter q for points where 9. If the eccentric angles of points P and Q on the ellipse
be q and p + q and a be the angle between the nor-
the line bx = ay intersects the ellipse x2 + y2 = 1? 2
a2 b2 mals at P and Q, prove that the eccentricity e is given
30. Prove that the sum of the squares of the reciprocals of
two perpendicular radius vectors of an ellipse is con- 2 1 — e2
by tan a = e2 (sin2 2q ) .
stant.
31. Find the eccentricity of the ellipse
4(x – 2y + 1)2 + 9(2x + y + 2)2 = 25. 10. The tangent and the normal at any point P of an el-
32. If two vertices of a rectangle lie on y = 2x + c and other lipse x2 + y2 =1 cut its major axis in points T and T¢
a2 b2
two vertices are (0, 4) and (–1, 2). Find c and other two
vertices such that the area of the ellipse inscribed in the respectively such that TT¢ = a. Prove that the eccentric
rectangle is 5p . angle of the point P is given by c2 cos2 j + cos j – 1
2
= 0.
LEVEL IV 11. A variable point P on an ellipse of eccentricity e is
(Tougher Problems for JEE joined to its foci S and S¢. Prove that the locus of the
Advanced) in-centre of the DPSS¢ is an ellipse whose eccentricity
is 2e .
1+ e
1. Find the maximum area of an isosceles triangle in- 12. The eccentric angle of any point P measured from the
semi-major axis CA is f. If S be the focus nearest to A
scribed in the ellipse x2 + y2 =1 with its vertex at one and –ASP = q, prove that
a2 b2
end of the major axis. [Roorkee, 1994] Ê qˆ 1+ e Ê jˆ
ËÁ 2¯˜ 1- e ËÁ 2 ˜¯
2. If a tangent drawn at a point (t2, 2t) on the parabola tan = tan
y2 = 4x is the same as the normal drawn at a point
13. Prove that the locus of the centroid of an equilateral
( 5 cos j, 2 sin j) on the ellipse 4x2 + 5y2 = 20. Find
x2 y2
the values of t = j. inscribed in an ellipse a2 + b2 = 1 is
3. Find the equation of the largest circle with centre (1, 0) (a2 + 3b2 )2 x2 + (3a2 + b2 )2 y2 =1
that can be inscribed in the ellipse x2 + 4y2 = 16. a2 (a2 - b2)2 a2(a2 - b2)2
[Roorkee, 1999]
4. Find the condition so that the line px + qy = r inter- 14. The tangent at a point P(a cos j, b sin j) of an ellipse
sects the ellipse x2 + y2 =1 in points whose eccentric x2 y2
a2 b2 a2 b2
+ =1 meets its auxiliary circle in two points,
angles differ by p . [Roorkee, 2001] the chord joining which subtends a right angle at the
4
centre. Prove that the eccentricity of the ellipse is
5. A point moves so that the sum of the squares of its dis-
tances from two intersecting straight lines is constant. 1.
Prove that its locus is an ellipse. 1 + sin2 q
6. Prove that the line joining the extremities of any pair of 15. Prove that the locus of the point of intersection of tan-
gents to an ellipse at two points whose eccentric angles
diameters of an ellipse which are at right angles, will differ by a constant a is an ellipse
touch a fixed circle.
7. If P be any point on the ellipse x2 + y2 =1 whose or- x2 + y2 = sec2 Ê a ˆ
a2 b2 a2 b2 ËÁ 2 ˜¯
dinate is y¢, prove that the angle between the tangent at 16. If two concentric ellipses be such that the foci of
one be on the other end and e and e¢ be their eccen-
P and the focal distance of P is tan —1 Ê b2 ˆ . tricities. Prove that the angle between their axes is
ËÁ aey¢ ˜¯
8. Prove that the ellipse x2 + y2 =1 and the circle x2 + y2 Ê e2 + e¢2 - 1ˆ
a2 b2 cos-1 Á ˜.
Ë ee¢ ¯
-1 Ê a -b ˆ
= ab intersect at an angle tan ËÁ ab ˜¯ .
5.20 Coordinate Geometry Booster
17. If the normals at the four points (x1, y1), (x2, y2), (x3y3) 8. If F1 and F2 be the feet of perpendiculars from the foci
and (x4, y4) on the ellipse x2 + y2 =1 are concurrent, S1 and S2 of an ellipse x2 + y2 = 1 on the tangent at any
prove that a2 b2 5 3
Ê 1 1 1 1ˆ point P on the ellipse, find the value of (S1F1) ◊ (S1F2).
ËÁ x1 x2 x3 x4 ¯˜
(x1 + x2 + x3 + x4 ) + + + =1 9. Find the minimum length of the intercept of any tan-
18. If q be the difference of the eccentric angles of two gent to the ellipse x2 + y2 = 1 between the co-ordi-
16 9
points on an ellipse, the tangent at which are at right nate axes.
angles. Prove that ab sin q = d1d2, where d1, d2 are the 10. Find the number of distinct normals that can be drawn
semi-diameters parallel to the tangents at the points and to the ellipse x2 + y2 = 1 from the point (0, 6).
169 25
a, b are the semi-axes of the ellipse.
19. From any point on the conic x2 + y2 = 4 , tangents are
a2 b2
x2 y2 Comprehensive Link Passage
a2 b2
drawn to the conic + =1. Passage I
Prove that the normals at the points of contact meet on Let C: x2 + y2 = r2 and E: x2 + y2 =1 intersect at four dis-
the conic a2x2 + b2 y2 = 1 (a2 - b2 )2 . 16 9
4 tinct points A, B, C, D. Their common tangents form a paral-
lelogram EFGH.
20. Show that the locus of the centre of the circle which
x2 y2 1. If ABCD is a square, then r is
a2 b2
cuts the ellipse + =1 in a fixed point (h, k) and (a) 12 2 (b) 12 12
55 (c)
(d) none
55
two other points at the extremities of a diameter is
2(a2x2 + b2y2) = (a2 – b2)(hx – ky). 2. If EFGH is a square, then r is
(a) 20 (b) 12 (c) 15 (d) none
Integer Type Questions 3. If EFGH is a square, the ratio of
1. Find the minimum area of the triangle formed by the ar (Circle C) is
tangent to the ellipse x2 + y2 = 1 . ar (circumcircle D EFG)
94
(a) 9 (b) 3 (c) 1 (d) none
2. Let P be a variable point on the ellipse x2 + y2 = 1 16 4 2
51
Passage II
with foci F1 and F2. If A be the area of the triangle A curve is represented by C:
DPF1F2, find the maximum value of A.
21x2 – 6xy + 29y2 + 6x – 58y – 151 = 0
Ê 16 j ˜¯ˆ 1. The eccentricity of the given curve is
ÁË 11
3. If the tangent at 4 cos j , sin to the (a) 1 (b) 1 (c) 2 (d) 2
3 3 3 5
ellipse 16x2 + 11y2 = 256 is also a tangent to the circle
2. The length of axes are
x2 + y2 – 2x – 15 = 0, find the number of values of f.
(a) 6, 2 6 (b) 5, 2 5 (c) 4, 4 5 (d) None
4. Find the area of a parallelogram formed by the tangents
at the extremities of a pair of conjugate diameters of an 3. The centre of the conic C is
ellipse x2 + 16y2 = 1 . (a) (1, 0) (b) (0, 0) (c) (0, 1) (d) None.
9
Passage III x2 + y2 = 1 whose ec-
5. Find the number of integral values of a for which the Let F and F¢ be the foci of the ellipse a2 b2
equation x2 + y2 = 1 represents an ellipse.
a - 10 4 - a centricity is e, P is a variable point on the ellipse.
1. The locus of the incentre of the DPFF¢ is a/an
6. Find the area of the greatest rectangle that can be in- (a) ellipse (b) hyperbola
scribed in an ellipse x2 + 4y2 = 4.
(c) parabola (d) circle
7. If P(x, y), F1 = ( 7, 0), F2 = (- 7, 0) and 9x2 + 16x2 2. The eccentricity of the locus of P is
= 144, find the value of (PF1 + PF2 – 2). (a) 2e (b) 2e (c) 1 (d) none
1-e 1+ e
Ellipse 5.21
3. The maximum area of the rectangle inscribed in the el- Passage VI
Tangents are drawn from the point P(3, 4) to the ellipse
lipse is x2 + y2 = 1 touching the ellipse at points A and B.
94
2abe2 2abe abe
(a) (b) (c) (d) none
1+ e 1+ e 1+ e
Passage IV 1. The co-ordinates of A and B are
x2 y2 (a) (3, 0) and (0, 2)
a2 b2
The ellipse + = 1 is such that it has the least area but (b) Ê - 8 , 2 161ˆ and Ê - 9, 8ˆ
ÁË 5 15 ¯˜ ÁË 5 5¯˜
contains the circle (x– 1)2 + y2 = 1.
Ê 161ˆ
1. The eccentricity of the ellipse is (c) ËÁ - 8 , 2 15 ˜¯ and (0, 2)
5
(a) 2 (b) 1 (c) 1
3 3 2 (d) none (d) (3, 0) and Ê - 9 , 8ˆ
ËÁ 5 5˜¯
2. The equation of the auxiliary circle of the ellipse is
2. The orthocentre of the DPAB is
(a) x2 + y2 = 13
2 (b) x2 + y2 = 5 (a) ÁÊË 5, 8 ˜ˆ¯ (b) ËÊÁ 7 , 285˜ˆ¯
(d) none 7 5
(c) x2 + y2 = 9
2 (c) ÁËÊ 11 , 85¯ˆ˜ (d) ÁÊË 8 , 75 ˜ˆ¯
5 25
3. The length of the latus rectum of the ellipse is
(a) 2 (b) 1 (c) 3 (d) 5/2 3. The equation of the locus of the point whose distances
Passage V from the point P and the line AB are equal is
The conic section is the locus of a point which moves in a (a) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0
plane in such a way that, the ratio of its distance from a fixed (b) 9x2 + y2 + 6xy – 54x + 62y + 241 = 0
point to a fixed straight line is constant. The fixed point is (c) 9x2 + 9y2 – 6xy – 54x – 62y + 241 = 0
called focus and the fixed straight line is called the directrix. (d) x2 + y2 – 2xy – 27x – 31y + 120 = 0
The constant ratio is called the eccentricity. It is denoted by e.
Matrix Match
If e is less than 1, the conic section is called an ellipse. A (For JEE-Advanced Examination Only)
line joining two points on the ellipse is called the chord of
the ellipse. If through a point C, any chord of an ellipse is 1. Match the following columns:
bisected, the point C is called the centre of the ellipse. Let the equation of the curve is x2 + 2y2 + 4x + 12y = 0.
Let the equation of the curve is Column I Column II
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 (A) The centre is (P) (- 2 - 11, -)
It will represents an ellipse if h2 < ab and D π 0, where (B) The focus is (Q) (–2, –3)
ahg
D= h b f (C) One extremity of the (R) ( 11 - 2, - 3)
gf c major axis is
(D) The latus rectum is (S) Ê 11 ˆ
ËÁ - 2 + 2 2˜¯
1. The length of the longest chord of the ellipse x2 + xy + , -
y2 = 1 is
(a) 2 (b) 1 (c) 2 2 (d) 1 (T) 22
2
2. The length of the chord passing through the centre 2. Match the following columns:
Let the equation of the ellipse is (x - 3)2 + ( y - 2)2 = 1.
and perpendicular to the longest chord of the ellipse 25 16
x2 + xy + y2 = 1 is
(a) 1 (b) 3 (c) 2 2 1 Column I Column II
2 2 3 (d) (A) The centre is (P) (8, 2)
(B) One extremity of major axis is (Q) (3, 2)
3 (C) One of the foci is (R) (6, 2)
(D) One extremity of minor axis is (S) (3, 6)
3. There are exactly n integral values of l for which the (E) The length of latus rectum is (T) 10
equation x2 + lxy + y2 = 1 represents an ellipse, then n (F) The focal distance is (U) 6.4
must be
(a) 0 (b) 1 (c) 2 (d) 3
4. The centre of the ellipse x2 + xy + 2y2 = 1 is
(a) (0, 0) (b) (1, 1) (c) (1, 2) (d) (2, 1)
5.22 Coordinate Geometry Booster
3. Match the following columns: (B) The length of PQ is (Q) 12 3
(R) 9 3
For all real p, the line 2px + 2 1 - p2 = 1 touches a (C) Area of DTPQ is (S) x + 2y = 2
fixed ellipse. (T) 2x + y = 1
(D) Area of quadrilateral
Column I Column II OPTQ is
(A) The eccentricity is (P) 1/2
(B) The latus rectum is (Q) 3/2
(C) The focal distance is (R) 2 Questions asked in Previous Years’
JEE-Advanced Examinations
(D) One extremity of major axis (S) (0, 1)
is
(E) One of the foci is (T) (0, 3/2) 1. The locus of a point whose distance from (–2, 0) is 2/3
4. Match the following columns: times its distance from the line x = - 9 is a/an
2
A tangent having slope –4/3 touches the ellipse
x2 + y2 = 1 at point P and intersects the major and (a) ellipse (b) parabola
18 32
minor axes at A and B respectively, where O is the cen- (c) hyperbola (d) none [IIT-JEE, 1994]
tre of the ellipse.
2. Let E be the ellipse x2 + y2 = 1 and C be the circle
94
x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1)
Column I Column II respectively, then
(a) Q lies inside C but outside E
(A) The distance between the parallel (P) 24 (b) Q lies outside both C and E
tangents having slopes –4/3 is
(c) P lies inside both C and E
(B) Area of DAOB is (Q) 7/24 (d) P lies inside C but outside E [IIT-JEE, 1994]
(C) If the tangent in first quadrant (R) 48/5 3. Let P be a variable point on the ellipse x2 + y2 =1
touches the ellipse at (h, k), the a2 b2
value of hk is
with foci F1 and F2. If A is the area of DPF1F2, find the
(D) If the equation of the tangent in- (S) 12 maximum value of A. [IIT-JEE, 1994]
tersecting positive axes is lx + my
= 1, the value of l + m is 4. The radius of the circle passing through the foci of the
5. Match the following columns: ellipse x2 + y2 = 1 , and having its centre at (0, 3) is
16 9
Let M(t, t + 1) is a point moving in a straight line and (c) 1 (d) 7
C1, C2, C3 be three conics whose equations are x2 + y2 = (a) 4 (b) 3 2 2
2, y2 = 8x and x2 + 2y2 = 1. Let M lies within the interior
[IIT-JEE, 1995]
of C1, C2, C3.
5. Let d be the perpendicular distance from the centre of
Column I Column II the ellipse x2 + y2 = 1 to the tangent drawn at a point
(A) C1 is a2 b2
(B) C2 is
(P) (3 - 2 2, 3 + 2 2) P on the ellipse.
(C) C3 is
(Q) ËÁÊ -1, - 13˜ˆ¯ If F1 and F2 are two foci of the ellipse, show that
(PF1 - PF2 )2 = 4a2 Ê - b2 ˆ . [IIT-JEE, 1995]
ËÁ1 d2 ˜¯
(R) Ê -1 - 3 , -1 + 3ˆ No questions asked in 1996.
ÁË 2 2 ˜¯
6. A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse
6. Match the following columns: x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and
Tangents are drawn from (2, 3) to the ellipse Q of the ellipse x2 + 2y2 = 6 are at right angles.
x2 + y2 = 1.
43 [IIT-JEE, 1997]
7. An ellipse has OB as a semi-minor axis. F and F¢ are
its foci, and the ellipse FBF¢ is a right angle. Find the
Column I Column II eccentricity of the ellipse. [IIT-JEE, 1997]
(A) The equation of PQ is (P) 3 ¥ 3
8. If P = (x, y) be any point on 16x2 + 25y2 = 400 with foci
2
F1 = (3, 0) and F2 = (–3, 0), then PF + PF2 is
(a) 8 (b) 6 (c) 10 (d) 12
[IIT-JEE, 1998]
Ellipse 5.23
9. Find the co-ordinates of all points P on the ellipse (a) 1 (a2 + b2 ) s.u. (b) 1 (a2 + b2 + ab) s.u.
2 3
x2 + y2 = 1, for which the area of DPON is maxi-
a2 b2 (c) 1 (a + b)2s.u. (d) ab s.u.
2 [IIT-JEE, 2005]
mum, where O denotes the origin and N, the foot of the
perpendicular from O to the tangent at P.
[IIT-JEE, 1999] No questions asked in between 2006-2007.
10. On the ellipse 4x2 + 9y2 = 1, the points at which the
17. Let P(x1, y1), Q(x2, y2), y1 < 0, y2 < 0, be the end-points
tangents are parallel to the line 9y = 8x are of the latus rectum of the ellipse x2 + 4y2 = 4. The equa-
tion of the parabola with latus rectum PQ are
(a) ÊÁË 2 , 15ˆ¯˜ (b) ÁÊË - 2 , 15ˆ˜¯
5 5 (a) x2 + 2 3y = 3 + 3 (b) x2 - 2 3y = 3 + 3
(c) ÁÊË - 2 , - 15¯˜ˆ (d) ÁËÊ 2 , - 15ˆ¯˜ (c) x2 + 2 3y = 3 - 3 (d) x2 - 2 3y = 3 - 3
5 5
[IIT-JEE, 2008]
[IIT-JEE, 1999]
18. The line passing through the extremity A of the major
x2 y2
11. Let P be a point on the ellipse a2 + b2 =1 0 < b < a. axis and extremity B of the minor axis of the ellipse
x2 + 9y2 = 9 meets its auxiliary circle at the point M.
Let the line parallel to y-axis passing through P meet Then the area of the triangle with vertices at A, M and
the circle x2 + y2 = a2 at the point Q such that P and the origin O is
Q are on the same side of x-axis. For two positive real (a) 31/10 (b) 29/10
numbers r and s, find the locus of the point R on PQ (c) 21/10 (d) 27/10
such that PR : RQ = r : s and p varies over the two
[IIT-JEE, 2009]
ellipses. [IIT-JEE, 2001] 19. The normal at P on the ellipse x2 + 4y2 = 16 meets the
12. Prove that, in an ellipse, the perpendicular from a focus x-axis at Q. If M is the mid-point of the line segment
upon any tangent and the line joining the centre of the PQ, the locus of M intersects the latus rectums of the
ellipse to the point of contact meet on the correspond- given ellipse at the points
ing directrix. [IIT-JEE, 2002] (a) (± 3 5 /2, ± 2/7) (b) (± 3 5 /2, ± 19 /7)
13. The area of the quadrilateral formed by the tangents at (c) (± 2 3, ±1/7) (d) (± 2 3, ± 4 3/7)
[IIT-JEE, 2009]
the end-points of latus rectum to the ellipse x2 + y2 = 1,
is 9 5 Comprehension
(a) 27/4 s.u. (b) 9 s.u.
(c) 27/2 s.u. (d) 27 s.u. Tangents are drawn from the point P(3, 4) to the ellipse
x2 + y2 = 1 touching the ellipse at points A and B.
[IIT-JEE, 2003] 94
14. If tangents are drawn to the ellipse x2 + 2y2 = 2, the
locus of the mid-point of the intercept made by the tan-
gents between the co-ordinate axes is 20. The co-ordinates of A and B are
(a) (3, 0) and (0, 2)
1 1 1 1
(a) 2x2 + 4y2 =1 (b) 4x2 + 2y2 =1 Ê 8, 2 16 ˆ ÊËÁ 9 8 ˆ˜¯
ËÁ 5 15 ˜¯ 5 5
(b) - and - ,
(c) x2 + y2 = 1 (d) x2 + y2 = 1 (c) Ê - 8, 2 16 ˆ and (0, 2)
24 42 ËÁ 5 15 ¯˜
[IIT-JEE, 2004] 9 85¯ˆ˜
5
15. Find the equation of the common tangent in first quad- (d) (3, 0) and ËÊÁ - ,
rant to the circle x2 + y2 = 16 and the ellipse x2 + y2 = 1. 21. The orthocentre of DPAB is
25 4
8 Ê 7 25ˆ
Also find the length of the intercept of the tangent be- (a) ÁÊË 5, 7 ˜ˆ¯ (b) ÁË 5 , 8 ˜¯
tween the co-ordinate axes. [IIT-JEE, 2005] Ê 11 8ˆ Ê 8 7ˆ
ÁË 5 5¯˜ ÁË 25 5 ¯˜
16. A triangle is formed by a tangent to the ellipse (c) , (d) ,
x2 + y2 =1 and the co-ordinate axes. The area of the [IIT-JEE, 2010]
a2 b2
22. The equation of the locus of the point whose distances
triangle cannot be less than
from the point P and the line AB are equal, is
5.24 Coordinate Geometry Booster
(a) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0 The m1 is the slope of T1 and m2 is the slope of T2, then
(b) x2 + 9y2 + 6xy – 54x + 62y – 241 = 0
(c) 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0 the value of Ê1 + m22 ˆ is... [IIT-JEE-2015]
(d) x2 + y2 – 2xy + 27x + 31y – 120 = 0. ÁË m12 ˜¯
[IIT-JEE, 2011] 27. Let E1 and E2 be two ellipses whose centers are at the
23. The ellipse E1 : x2 + y2 =1 is inscribed in a rectangle origin. The major axes of E1 and E2 lie along the x-axis
9 4 and the y-axis, respectively. Let S be the circle x2 + (y
– 1)2 = 2. The straight line x + y = 3 touches the curves
R whose sides are parallel to the co-ordinate axes.
S, E1 ad E2 at P, Q and R, respectively. Suppose that
Another ellipse E2: passing through the point (0, 4)
circumscribes the rectangle R. The eccentricity of the PQ = PR = 2 2 . If e1 and e2 are the eccentricities of
3
ellipse E2 is
E1 and E2, respectively, then the correct expression(s)
(a) 2 (b) 3 (c) 1 (d) 3 is(are)
2 2 2 4
(a) e12 + e22 = 43 (b) e1e2 = 2 7
[IIT-JEE, 2012] 40 10
24. A vertical line passing through the point (h, 0) inter- (c) | e12 - e22 | = 5 (d) e1e2 = 3
8 4
sects the ellipse x2 + y2 = 1 at the points P and Q. Let
43 [IIT-JEE-2015]
the tangents to the ellipse at P and Q meet at R. 28. Comprehension
If D(h) = area of DPQR, Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the
foci of the ellipse x2 + y2 = 1
D1 = max D(h)
1 98
2 £ h £1 Suppose a parabola having vertex at the origin and fo-
and D2 = 1min D(h)
2
£ h £1 cus at F2 intersects the ellipse at point M in the first
quadrant and at point N in the fourth quadrant.
then 8 D1 - 8D 2 =........
5 (i) The orthocentre of the triangle F1MN is
[IIT-JEE, 2013]
9 2
25. If the normal from the point P(h, 1) on the ellipse (a) Ê - 10 , 0¯ˆ˜ (b) Ê 3 , 0˜¯ˆ
x2 + y2 = 1 is perpendicular to the line x + y = 3, then ÁË ÁË
63
(c) Ê 9 , 0ˆ˜¯ (d) Ê 2 , 6¯ˆ˜
ÁË 10 ËÁ 3
the value of h is... [IIT-JEE, 2014]
26. Suppose that the foci of the ellipse x2 + y2 =1 are (f1, (ii) If the tangents to the ellipse at M and N meet at R
9 5 and the normal to the parabola at M meets the x-
axis at Q, then the ratio of area of the triangle MQR
0) where (f2, 0). Let P1 and P2 be two parabolas with a to area of the quadrilateral MF1NF2 is
common vertex at (0, 0) and with foci at (f1, 0) and (2f2, (a) 3:4 (b) 4:5 (c) 5:8 (d) 2:3
0), respectively. Let T1 be a tangent to P1 which passes
through (2f2, 0) and T2 be a tangent to P2 which passes [IIT-JEE-2016]
through (f1, 0).
LEVEL I ANSWERS
2. 10 6. e = 5 - 1
3. 4 < a < 10 2
4. 1
7. 0
3 8. q = p
5. x2 + y2 = 1
6
16 12 9. 24
10. (x - 3)2 + ( y — 2)2 = 1
16 9
Ellipse 5.25
11. SQ = 40 64. Ê - a2l , - b2m ˆ
3 ÁË n n ¯˜
12. (9p – 6) s.u. 66. (1, 1)
13. x2 + y2 = 1
67. a2c4 x2 + b2d4 y2 = 1
43
x2 y2 1
14. 2013 68. a4 + b4 = c2
15. 2
16. y = - 1 x + 13 69. a6 + b6 = (a2 - b2)2
x2 y2
2
17. ÁËÊ ± (- 48) , ±536ˆ˜¯ 70. 4a2x2 = y2b2 + 4a2a2
5
71. y = - b2x
18. l and ±5 a2m
19. y = 2x ± 14 72. P(a cos j, b sin j),
34. x2 + y2 = 25 Q(–a cos j, –b sin j),
35. x2 + y2 = 25
36. 4x2 + 3y2 – 12xy + 4x + 6y – 3 = 0 S(–a sin j, b cos j),
and R(a sin j, –b cos j)
cosec a sec a 1 75 x2 + y2 = 2
50. cosec b sec b 1 = 0 a2 b2
cosec g sec g 1 76. x2 + y2 = 1
a2 b2 2
Ê x2 y2 ˆ 1 1
51. ÁË a4 + b4 ˜¯ = a2 + b2 77. a3y + b3x = 0
52. x2 + y2 = 1 78. a2 + b2 = 2
a4 b4 c2 c2 d2
Ê x2 y2 ˆ 2 79. 2(a2x2 + b2y2)3 = (a2 – b2)(a2x2 – b2y2)2
ÁË a2 b2 ˜¯
53. (a2 + b2 ) + = (x2 + y2) 81. x2 + y2 = 2
a2 b2
Ê x2 y2 ˆ 2 Ê c2 ˆ
ÁË a2 b2 ˜¯ ÁË x2 + y2 ˜¯ 1
54. + = 82.
3
55. x ◊ y = a2b2 83. 4x + 3y = 12
l 84. x + y = 2
Ê a6 b6 ˆ Ê x2 y2 ˆ 2
ËÁ x2 y2 ˜¯ ËÁ a2 b2 ˜¯
56. + + = (a2 - b2)2 LEVEL II
57. a 2 Ê x2 + y2 ˆ = Ê x2 + y2 ˆ 2 1. (c) 2. (c) 3. (a) 4. (c) 5. (a)
ËÁ a4 b4 ˜¯ ÁË a2 b2 ˜¯ 6. (c) 7. (a) 8. (a) 9. (b) 10. (c)
11. (d) 12. (d) 13. (a) 14. (a) 15. (b)
58. Ê x2 + y2 ˆ 2 Ê 1 + 1ˆ = Ê x2 + y2 ˆ 16. (a, b) 17. (d) 18. (a) 19. (a) 20. (a,b,c,d)
ÁË a2 b2 ˜¯ ÁË a2 b2 ˜¯ ÁË a4 b4 ˜¯ 21. (a) 22. (c) 23. (a) 24. (a) 25. (d)
26. (c) 27. (b) 28. (c) 29. (c) 30. (b)
59. Ê x2 + y2 ˆ = 1 31. (a) 32. (d) 33. (a) 34. (b) 35. (d)
ËÁ a2 b2 ˜¯ 2 36. (c) 37. (c) 38. (d) 39. (d) 40. (a)
41. (a) 42. (c) 43. (a) 44. (a) 45. (d)
60. Ê x2 + y2 ˆ = Ê x2 + y2 ˆ 2 46. (c) 47. (d) 48. (a) 49. (a) 50. (d)
ËÁ a2 + b2 ¯˜ ÁË a2 b2 ¯˜ 51. (b) 52. (c) 53. (c) 54. (c) 55. (a)
56. (b) 57. (b) 58. (a) 59. (b) 60. (c)
61. b2x2 + a2y2 = ab2 xc 61. (a) 62. (c) 63. (d) 64. (c) 65. (c)
66. (d) 67. (d) 68. (d) 69. (c) 70. (b)
62. a 2 x2 + b 2 y2 = Ê x2 + y2 ˆ 2
a4 b4 ÁË a2 b2 ¯˜
63. x= a
e
5.26 Coordinate Geometry Booster
LEVEL III 30. 1 + 1 = 1 + 1
r12 r22 a2 b2
1. b (a2 - b2 ) 31. 5
3
2. 4
32. c = 14 or –6, A = (–4, 6), D = (3, 0)
4. j = p - tan2 2, t = - 1 ,j = p + tan -1 (2) ,
5
t = 1 ;j = ±p ,t = 0 LEVEL IV
52
5. 1 1. 3 3 ab
2 4
7. Ê a2 , b2 ˆ ËÊÁ 3 1¯˜ˆ
Á (a2 + b2) ˜ 2
Ë (a2 + b2 ) ¯ 2. 1 s.u., ,
8. (x - 1)2 + y2 = 121 p 3. (x - 1)2 + y2 = 11
9 8 3
9. a 2 p2 + b2q2 = r 2sec2 Ê ˆ ÊÁË p ¯˜ˆ
ËÁ ¯˜ 8
4. r2 = cos2 ¥ (a2 p2 + b2q2 )
12. 27 s.u.
13. p INTEGER TYPE QUESTIONS
6
1. 6 s.u.
14. 1 + 1 =1 2. 2 s.u.
2x2 4y2 3. 2
4. 3 s.u.
15. ab s.u. 5. 5
6. 4 s.u.
16. Y = — 2 x + 4 7 ; 14 7. 6
3 33 8. 3
9. 7
17. x2 + y2 = 1 10. 3
19. 4
20. 2
3- 5
COMPREHENSIVE LINK PASSAGES
21. Ê b2x2 + a2 y2 ˆ = 1 Ê x2 - y2 ˆ Passage I.: 1(a) 2(d) 3(c)
ËÁ a4 y2 + b4 x2 ˜¯ c2 ËÁ1 - a2 b2 ¯˜ Passage II: 1(b) 2(a) 3(c)
Passage III: 1(a) 2(b) 3(a)
22. y2 = Ê 2b2 ˆ x Passage IV: 1(a) 2(c) 3(b) 4(a)
ÁË a ˜¯ Passage V: 1(c) 2(c) 3(d) 4(a)
Passage VI: 1(a) 2(c) 3(a)
23. (x2 + y2)(x2 + y2 – a2 – b2 = 2xy(a2 – b2)
24. {(x2 – y2) + (b2 + a2)} tan (2a) + 2xy MATRIX MATCH
25. 5 2 1. (A) Æ (Q); (B) Æ (P, R); (C) Æ (S); (D) Æ (T)
2. (A) Æ (Q); (B) Æ (P); (C) Æ (R); (D) Æ (S);
26. x2 + 9y2 = 1
94 (E) Æ (U); (F) Æ (T)
3. (A) Æ (Q); (B) Æ (P); (C) Æ (R); (D) Æ (S);
27.
28. (E) Æ (T)
29. p or 5p 4. (A) Æ (R); (B) Æ (P); (C) Æ (S); (D) Æ (Q)
5. (A) Æ (R); (B) Æ (Q); (C) Æ (P)
44 6. (A) Æ (S); (B) Æ (P); (C) Æ (R); (D) Æ (Q)
Ellipse 5.27
HINTS AND SOLUTIONS
LEVEL I (c) Co-vertices: (0, ±b)
fi X = 0, Y = ±3
1. (i) The given equation of the ellipse is 9x2 + 16y2 = 144 fi x – 1 = 0, y – 2 = ±3 =
fi x = 1, y = 2 ± 3
fi x2 + y2 = 1 fi x = 1, y = 5, –1
16 9 Hence, the co-vertices are (1, 5), (1, –1).
(a) Centre: C(0, 0) (d) The length of the major axis = 2a = 6
(b) Vertices: A(a, 0) A¢(–a, 0) = A(4, 0) and A¢(– (e) The length of the minor axis = 2b = 4
4, 0) (f) The length of the latus rectum
(c) Co-vertices: B = (0, b) and B¢ =(0, –b)
2b2 = 18 = 9
fi B = (0, 3) and B¢ = (0, –3) a 42
(d) Length of the major axis: 2a = 8
(e) Length of the minor axis: 2b = 6 (g) Eccentricity,
b2 1- 9 = 7 e= 1- b2 = 1- 2 = 1
a2 16 4 a2 33
(f) Eccentricity = e = 1- =
(h) Equations of the directrices:
(g) Lengths of the latus rectum = 2b2 = 18 = 9 X = ± a = ± 3 = ±3 3
a 42 e 1/ 3
(h) Equations of the directrices: fi x =1±3 3
x=±a (i) End-points of the latus recta:
e
L Ê ae, b2 ˆ ; L¢ Ê ae, - b2 ˆ
fi x = ± 4 = ± 16 ËÁ a ˜¯ ÁË a ˜¯
77
4 = L Ê 3, 9ˆ ; L¢ Ê 3, - 9 ˆ
ÁË 2¯˜ ÁË 2 ˜¯
(i) End-points of the latus recta:
L Ê ae, b2 ˆ , L¢ Ê ae, — b2 ˆ 2. The given equation of the ellipse is
ËÁ a ¯˜ ÁË a ¯˜ 16x2 + 25y2 = 400
= L ÊÁË 7, 94ˆ˜¯ , L¢ ÊËÁ 7, - 94ˆ¯˜ fi x2 + y2 = 1
25 16
Thus, the sum of the focal distances
(ii) The given equation of the ellipse is = 2a = 10 units x2 + y2 =1
2x2 + 3y2 – 4x – 12y + 8 = 0 3. The given equation of an ellipse is
10 - a a — 4
fi 2(x2 – 2x + 1) + 3(y2 – 4y + 4) = 6
fi 2(x – 1)2 + 3(y – 2)2 = 6 fi (a – 4)x2 + (10 – a)y2 – (a – 4)(10 – a) = 0
fi (x — 1)2 + ( y — 2)2 = 1 Since, the given equation represents an ellipse,
32 h2 – ab < 0
fi 0 – (a – 4)(10 – a) < 0
fi (a – 4)(a – 10) < 0
fi X 2 + Y 2 = 1 , where X = x – 1, Y = y – 2 fi 4 < a < 10
32
Hence, the length of the interval = 10 – 4 = 6.
Thus, the value of m is 6.
(a) Centre: (0, 0) 4. Let the foci of the ellipse are S and S¢ respectively.
fi X = 0, Y = 0
fi x – 1 = 0 and y – 2 = 0 Then S = (5, 12) and S¢ = (24, 7).
fi x = 1 and y = 2
Hence, the centre is (1, 2). Thus, the centre of the ellipse is C Ê 29 , 19ˆ .
ËÁ 2 2 ˜¯
(b) Vertices : (±a, 0)
fi X = ±a, Y = 0 Now OC = 1 ¥ 292 + 192 = 1 ¥ 1202
fi x – 1 = ±3, y – 2 = 0 22
fi x = 1 ± 3, y = 2
fi x = 4, –2, y = 2 We have,
Hence, the vertices are (4, 2) and (–2, 2)
SS¢ = 386
fi e = 386 = 386 = 386 = 1
2a 1202 1202 3
5.28 Coordinate Geometry Booster
5. Let S(2, 0) and S¢(–2, 0) are two foci of the ellipse and fi 4 cos2 q = 3
C(0, 0) be the centre of the ellipse. fi cos q = 3 = 3
42
We have,
fi q=p
SS¢ = 4 6
fi 2ae = 4 Hence, the eccentric angle is p .
6
fi 2a ¥ 1 = 4 fia=4
2
Also, b2 = a2 (1 - e2 ) = 16ÁÊË1 - 1ˆ = 12 9. Let Q be any point S P
4˜¯ on the given ellipse,
M
fi b = 12 = 2 3 whose co-ordinates Q
are (4 cos2 q, 3 sin2
Hence, the equation of the ellipse is N C
x2 + y2 =1 q).
a2 b2
Thus, PQ = 8 cos q R
x2 + y2 = 1 and QR = 6 sin q
16 12
fi Thus, the area of the rectangle PQRS
= PQ ¥ QR
6. Given 2ae = 2b2 = 48 sin q cos q
a
= 24 sin 2q
fi b2 = a2e
fi a2(1 – c2) = a2c Hence, the area of the greatest rectangle
fi (1 – c2) = c
fi c2 + c – 1 = 0 = 24 sq.u. at q = p
4
fi e= 5 -1 10. The co-ordinates of the given point are
2 x = 3 + 4 cos q, y = 2 + 3 sin q
Hence, the result x2 y2 fi (x - 3) = cos q, (y — 2) = sin q
The equation of the given ellipse is a2 b2 43
7. + =1 (x - 3)2 (y — 2)2 cos2q sin 2q
42 32
It is given that the length of the latus rectum = 1 ¥ the fi + = + =1
length of the major axis. 2 fi (x - 3)2 + ( y — 2)2 = 1
16 9
fi 2b2 = a
a which is the required locus.
fi b=a 11. The equation of the ellipse is
16x2 + 25y2 = 400
As we know that,
fi x2 + y2 = 1
eccentricity = 1- b2 = 1-1= 0 25 16
a2
As we know that, if SP and SQ are the focal segments
8. Let any point on the ellipse x2 + y2 = 1 is of a focal chord PSQ, then 1 + 1 = 1
62
SP SQ a
P( 6 cos q, 2 sin q) and C be the centre of the el-
B
lipse. P
B A¢ A
P(q) CS
A¢ C A B¢ Q
B¢ fi 1+ 1 =1
8 SQ 5
Therefore, 1 =1-1=8-5= 3
SQ 5 8 40 40
CP = 5 fi
fi 6 cos2 q + 2 sin2 q = 5
Ellipse 5.29
fi SQ = 40 3 = 2m + 16m2 + 9
3
fi (3 – 2m)2 = 16m2 + 9
Hence, the length of SQ is 40/3. fi 9 – 6m + 4m2 = 16m2 + 9
fi 12m2 + 6m = 0
12. As we know that the area of an ellipse x2 + y2 =1 is fi 2m2 + m = 0
pab. a2 b2 fi m(2m + 1) = 0
Y fi m = 0, — 1
2
B
C Hence, the equation of tangents are y = 3 and
y = - 1 x + 13.
X¢ O AX
2
Y¢ 17. The given line is
Thus, the area of OACBO 3x + 4y = 5
= Ê 3 ¥ p ◊ 4 ◊ 3 - 1 ◊ 4 ◊ 3˜¯ˆ fi 4y = –3x + 5
ËÁ 4 2
and the ellipse is
= (9p – 6) s.u. 9x2 + 16y2 = 144
N = x12 + y12 - 1
13. Let fi y = - 3 x + 5 and x2 + y2 = 1
43 4 4 16 9
Then, N = 4 + 9 - 1 = 1 + 3 - 1 = 3 As we know that the line y = mx + c will be a tangent
43
to the ellipse x2 + y2 =1, the points of contacts are
Since, the value of N is positive at (2, 3), so the point a2 b2
(2, 3) lies outside of the ellipse x2 + y2 = 1 .
Ê ± a2m , ± b2 ˆ
43 ÁË c c ˜¯
14. Since the point (l, –l) be an interior point of an ellipse
Ê ± 16ÊÁË - 3ˆ ˆ
4x2 + 5y2 = 1, then Á 5 4¯˜
4l2 + 5l2 – 1 < 0 = Á 4 , ±9˜
ÁË ˜
fi 9l2 – 1 < 0 5 ˜¯
4
fi (3l + 1)(3l – 1) < 0
= Ê ± (- 48) , ± 36ˆ
fi -1 <l < 1 ËÁ 5 5 ¯˜
33
Therefore, m = 2 18. The equation of the given ellipse is
3 9x2 + 16y2 = 144
fi (3m – 2)2013 + 2013 = 0 + 2013 = 2013 fi x2 + y2 = 1
16 9
15. We have,
4.4 + 3.3 – 12 = 16 + 9 – 12 = 13 > 0 The line y = x + l will be a tangent to the ellipse, if
c2 = a2m2 + b2
Thus, the point (2, 3) lies outside of the ellipse.
Thus, the number of tangents = 2 fi l2 = 16.1 + 9 = 25
16. The given ellipse is fi l = ±5
Hence, the values of l are ±5.
9x2 + 16y2 = 144 19. The given ellipse is x2 + y2 = 1
fi x2 + y2 = 1
34
16 9 The equation of any ellipse is
The equation of any tangent to the given ellipse is
y = mx + a2m2 + b2
fi y = mx + 16m2 + 9 y = mx ± a2m2 + b2
which is passing through (2, 3), so
fi y = 2x ± 3.4 + 2 = 2x ± 14
5.30 Coordinate Geometry Booster
20. The equation of any tangent to the ellipse x2 + y2 =1 fi Ê Ê r2 - b2 ˆ ˆ
at P(q) is a2 b2 q = tan-1 ÁÁË ÁË a2 - r2 ˜¯ ˜˜¯
x cos q + y sin q = 1 …(i) Hence, the result.
ab
22. Let the end-points of a latus rectum are
The equation of perpendiculars from centre (0, 0) to
L(ae, 0) and L (–ae, 0).
tangent is
The tangent at L is xe + y = 1 and the tangent at L¢ is
x sin q - y cos q = 0 …(ii) aa
ba xe - y = 1 .
From Eq. (ii), we get aa
sin q = cos q = 1 On solving, we get,
by ax a2x2 + b2 y2 …(iii) x = a and y = 0.
e
The locus of the feet of perpendiculars is the point of
fi x = a is the equation of directrix to the ellipse
intersection of (i) and (ii). e
It is obtained by eliminating q between Eqs (i) and (ii). x2 y2
a2 b2
Squaring Eqs (i) and (ii) and adding, we get + = 1.
x2 cos2q + y2 sin 2q + x2 sin 2q + y2 cos2q =1 23. The equation of any tangent to the ellipse at P(q) is
a2 b2 b2 a2
AB: x cos q + y sin q = 1.
fi Ê x2 + y2 ˆ cos2q + Ê y2 + x2 ˆ sin 2q =1 ab
ÁË a2 a2 ¯˜ ËÁ b2 b2 ˜¯
Thus A = (a sec q, 0) and A = (0, b cosec q)
fi (x2 + y 2 ) Ê cos2q + sin2q ˆ =1
ÁË a2 b2 ¯˜ Let (h, k) be the mid-point of the tangent AB.
fi (x2 + y2)(b2 cos2 q + a2 sin2 q) = a2b2 Therefore, h = a sec q and k = b cosec q
22
fi (x2 + y 2 ) Ê b2a2x2 + b2a2 y2 ˆ = a2b2
ÁË a2x2 + b2 y2 a2 x2 + b2 y2 ˜¯ fi cos q = a and sin q = b
2h 2k
fi (x2 + y2)2 = (a2x2 + b2y2) Now squaring and adding, we get
Now, put, x = r cos q and y = r sin q, we get a2 + b2 =1
r2 = a2cos2 q + b2 sin2 q. 4h2 4k 2
Hence, the result. fi a2 + b2 = 4
21. The equation of circle is x2 + y2 = r2. h2 k2
So the equation of any tangent to the ellipse is Hence the locus of (h, k) is
y = mx + a2m2 + b2 a2 + b2 = 4
x2 y2
If it is a tangent to a circle also, then the length of the
perpendicular from the centre (0, 0) of a circle is equal Now, put x = r cos q c y = r sin q , we get,
to the radius of a circle. 4r2 = a2 sin2 + b2 cos2 q ,
Thus, a2m2 + b2 = r Hence, the result.
1 + m2
24. Given ellipses are x2 + y2 =1 …(i)
a2 b2 …(ii)
and x2 + y2 = a + b
fi a2m2 + b2 = r2 (1 + m2 ) = r2 + r2m2 ab
fi (a2 – r2)m2 = (r2 – b2)
fi x2 + y2 = 1
fi m2 = Ê r2 - b2 ˆ a(a + b) b(a + b)
ËÁ a2 - r2 ¯˜
Let R(h, k) be the points of intersection of the tangents
fi m= Ê r2 - b2 ˆ to the ellipse (ii) at P and Q. Then PQ will be the chord
ËÁ a2 - r2 ˜¯ of contact.
Thus its equation will be
fi tan q = Ê r2 - b2 ˆ hx + ky = 1
ËÁ a2 - r2 ¯˜ a(a + b) b(a + b)
Ellipse 5.31
fi y = - bh x + b(a + b) …(iii) Therefore, the equation of a director circle is
ak k x2 + y2 = a2 + b2
Since the line (iii) is a tangent to the ellipse (i), so we This means that the centre of the ellipse will always
have
remain at a constant distance a2 + b2 from P.
b2(a + b)2 = a2 ¥ b2 ¥ h2 + b2 Hence, the locus of the centre is a circle.
k2 a2 k2
27. The equation of any tangent to the ellipse x2 + y2 =1
fi b2(a + b)2 = b2h2 + b2k2 at P(q) is a2 b2
fi h2 + k2 = (a + b)2 Y
Thus the locus of (h, k) is
x2 + y2 = (a + b)2 QP
which is the director circle of the ellipse p1 p2
X¢ S¢ O S
x2 + y2 = a + b X
ab
Hence, the tangents at P and Q are at right angles.
25. Let two foci of an ellipse are Y¢
S(ae, 0) and S(–ae, 0). x cos q + y sin q = 1 …(i)
ab
The equation of the tangent to the ellipse x2 + y2 =1
at P(q) is x cos q + y sin q = 1. a2 b2
The equation of perpendiculars from foci (±ae, 0) to
ab the tangent is
Let p1 and p2 be the lengths of perpendiculars from the x sin q - y cos q = ± ae sin q …(ii)
given foci to the given ellipse. ba b
Thus, p1 = 1 - e cos q Locus of the feet of perpendiculars is the point of inter-
section of (i) and (ii).
cos2q + sin 2q It is obtained by eliminating q between Eqs (i) and (ii).
a2 b2 Squaring Eqs (i) and (ii) and adding, we get
and p2 = 1 + e cos q Ê cos2q sin2q ˆ Ê cos2q sin2q ˆ
ËÁ a2 b2 ¯˜ ËÁ a2 b2 ¯˜
cos2q sin 2q fi x2 + + y 2 +
a2 b2
+
1 - e cos q 1 + e cos q =1+ a 2 e2sin 2q
b2
fi p1 ¥ p2 =
cos2q sin 2q cos2q sin 2q
a2 + b2 a2 + b2 fi (x2 + y 2 Ê cos2q + sin 2q ˆ = a2 Ê cos2q + sin 2q ˆ
)ËÁ a2 b2 ¯˜ ÁË a2 b2 ˜¯
= Ê 1 - e2cos2q ˆ fi (x2 + y2) = a2
Á ˜
Á cos2q + sin 2q ˜ Hence, the result.
Ë a2 b2 ¯ 28. The tangent at P(q) is x cos q + y sin q = 1.
= Ê a2b2 (1 - e2cos2q) ˆ ab
ÁË b2cos2q + a2sin2q ˜¯
Y
= Ê a2 a2b2 (1 - e2cos2q) 2q ˆ M P(q)
ÁË (1 - e2 ) cos2q + a2sin ˜¯ p
= b2 O S(æ, 0) X
26. Consider P is the point of intersection of two perpen-
dicular tangents. Thus the locus of P is the director
circle.
5.32 Coordinate Geometry Booster
Also SP = a – ex = a(1 – e cos q) fi a2b2 = a 2sin 2q + b2cos2q
p2
Since p is the length of perpendicular from the focus
S(ae, 0), = a2 (1 - cos2q) + b2 (1 - sin2q)
then, p = e cos q - 1 = a2 + b2 – (a2 cos2 q + b2 sin2 q)
= a2 + b2 – r2
Ê cos2q + sin2q ˆ
ËÁ a2 b2 ˜¯ Hence, the result.
fi 1 = (b2cos2q + a2sin2q ) 30. Any tangent to the ellipse
p2 a2b2 (e cos q - 1)2
x cos q + y sin q = 1
fi b2 = (b2cos2q + a2sin2q ) ab
p2 a2 (e cos q - 1)2 Let it meet the x-axis at A and y-axis at B. Then the co-
ordinates of A and B are
= È a2 (1 - e2) cos2q + a2sin2q ˘ A = Ê a q , 0˜¯ˆ and B = Ê 0, b q ˆ
Í a2 (e cos q - 1)2 ˙ ÁË cos ÁË sin ¯˜
Î ˚
Let M(h, k) be the mid-point of AB.
= È (1 - e2 ) cos2q + sin 2q ˘ Then 2h = a and 2k = b
Í (e cos q - 1)2 ˙ cos q sin q
Î ˚
fi h = a and k = b
= È (1 - e2cos2q) ˘ 2 cos q 2 sin q
Í (e cos q - 1)2 ˙
Î ˚ fi cos q = a and sin q = b
2h 2k
= È1 + e cos q ˘
ÍÎ1 - e cos q ˙ Squaring and adding, we get
˚
a2 b2
Also, 4h2 + 4k 2 =1
2a - 1 = 2a - 1 = 1 + e cos q
SP a(1 - ecos q ) 1 - e cos q fi a2 + b2 = 4
h2 k2
Hence, b2 = 2a - 1
p2 SP Hence, the locus of M(h, k) is
29. Let the point P be (a cos q c b sin q). a2 + b2 = 4
The equation of the tangent at P be x2 y2
x cos q + y sin q = 1 31. The tangent at P(a cos q, b sin q) is
ab
x cos q + y sin q = 1
ab
Let it meets the directrix x = a at Q, where Q is
e
MP Ê a b(e - cos q)ˆ
ÁË e , e sin q ¯˜ .
r
X¢ O X
Slope of SP = m1 = b(e - cos q )
e sin q
Y¢ Slope of SQ = m2 = b(e - cos q) Ê a - aeˆ¯˜
e sin q ËÁ e
0+ 0-1
Now, p =
cos2q sin 2q b(e - cos q )
a2 + b2 = a(1 - e2 )sin q
Thus, m1 ¥ m2
fi 1 = cos2q + sin 2q = b sin q e) ¥ b(e - cos q )
p2 a2 b2 a (cos q - a(1 - e2 )sin q
= a2sin2q + b2cos2q = - b2 e2 ) = - b2 = -1
a2b2 a2 (1 - b2
Hence, the result.
Ellipse 5.33
32. Any tangent to the ellipse is …(i) fi sin 2q = a2 and cos2q = b2
x cos q + y sin q = 1 a2 + b2 a2 + b2
ab
Its point of contact is P(a cos q b sin q) and its slope is Hence the point P is
- b cot q
Ê a2 , ± b2 ˆ
a Á± a2 + b2 ˜
Ë a 2 + b 2 ¯
Also the focus is S(ae, 0). 34. As we know that the locus of the point of intersection
Any line through the focus S and perpendicular to the of two perpendicular tangents is the director circle.
tangent (i) is Hence, the equation of the director circle to the given
y - 0 = a tan q (x - ae) …(ii) ellipse is
b
x2 + y2 = a2 + b2 = 16 + 9 = 25
Also the equation of CP is fi x2 + y2 = 25
y - 0 = a tan q (x - 0)
b …(iii) 35. As we know that the locus of the point of intersection
of two perpendicular tangents to an ellipse is the direc-
Eliminating q between Eqs (ii) and (iii), we get tor circle.
Ê a2 ˆ ÁËÊ x -x ae ˆ˜¯ = 1 Hence, the equation of the director circle is
ÁË b2 ¯˜ x2 + y2 = a2 + b2 = 4 + 1 = 5
fi x2 + y2 = 5
fi ÊËÁ x - ae ¯ˆ˜ = Ê b2 ˆ which is the required locus of P.
x ÁË a2 ¯˜
36. The equations of the pair of tangents to the ellipse
fi ËÊÁ1 - ae ˆ = Ê b2 ˆ 2x2 + 3y2 = 1 from the point (1, 1) is
x ¯˜ ÁË a2 ¯˜ (2x2 + 3y2 – 1)(2 + 3 – 1) = (2x + 3y – 1)2
fi 4(2x2 + 3y2 – 1) = (2x + 3y – 1)2
Ê b2 ˆ Ê ae ˆ fi 4(2x2 + 3y2 – 1) = 4x2 + 9y2 + 1 + 12xy – 4x – 6y
ÁË1 - a2 ¯˜ ÁË x ˜¯
fi = fi 4x2 – 3y2 – 12xy – 4x + 6y – 3 = 0
37. The equation of the given ellipse is
fi Ê a2 (1 - e2 ) ˆ = Ê ae ˆ 3x2 + 2y2 = 5
ËÁ1 - a2 ¯˜ ËÁ x ¯˜
fi x2 + y2 = 1
5/3 5/2
e2 ËÁÊ ae ¯˜ˆ
fi = x The equation of any tangents to the ellipse are
fi x=a y = mx ± b2m2 + a2
e
which is passing through (1, 2), so
Hence, the result.
33. Consider a point P on the ellipse whose coordinates are 2 = m ± 5 m2 + 5
23
(a cos q b sin q)
fi (2 — m)2 = 5 m2 + 5
The equation of the tangent at P is 23
x cos q + y sin q = 1
ab fi (2 — m)2 = 15m2 + 6
5
Since the tangent makes equal angles with the axes, so
its slope is ±1. fi 9m2 + 24m – 14 = 0
Let its roots are m1 and m2.
Thus, - b cos q = ±1 Then m1 + m2 = –24/9 and m1m2 = –14/9
a sin q Let q be the angle between them.
Then
fi b2cos2q =1
a 2sin 2q tan (q) = m2 - m1
1 + m1m2
fi sin 2q = cos2q = 1 b2
a2 b2 a2 + = (m1 + m2 )2 - 4m1m2
1 + m1m2
5.34 Coordinate Geometry Booster
64 + 56 120 ¥ 9 = 12 ¥ 5 B
35 6 P
= 99 =
1 - 14
9 A¢ CG A
F
G¢
Ê 12 6 ˆ
fi q = tan -1 ÁË 5 ˜¯
B¢
38. Do yourself. N
39. Do yourself. The equation of the normal at P is
ax sec q – by cosec q = a2 – b2
40. Do yourself. …(i)
41. Do yourself. b2 Since the line (i) meets the major axis at G and minor
a axis at G¢ respectively, then
42. Let one end of a latus rectum is L Ê ae, ˆ and minor
axis be B(0, –b). ÁË ˜¯
The equation of the normal to the ellipse at L is G = Ê a2 - b2 cos q, ˆ
ËÁ a 0¯˜
a x - ay = a2 - b2 and E = Ê 0, - a2 - b2 sin ˆ
e ÁË a q¯˜
which is passing through B¢(0, –b), so we have PF = CQ = length of perpendicular from C(0, 0) on the
ab = a2 – b2 tangent x cos q + y sin q = 1 at P.
fi ab = a2 – a2(1 – 32) = a2e2 ab
fi b = ae2 = ab
fi b2 = a2c4
fi a2(1 – e2) = a2e4 a2sin2q + b2cos2q
fi e4 + e2 – 1 = 0
Also, PG = b b2cos2q + a2sin2q
Hence, the result. a
43. Let the ellipse be x2 + y2 =1 and P(a cos q b sin q) be and PG¢ = a b2cos2q + a2sin2q
a2 b2 b
any point on the ellipse. Thus, PF ◊ PG = b2 and PF ◊ PG¢ = a2 .
Then, 45. The equation of the normal to the ellipse at P(q) is
SP = a – ex = a – ae cos q = a(1 – e cos q) ax sec q – by cosec q = a2 – b2)
Y
and P
SP = a + ex = a + ae cos q = a(1 + e cos q)
The equation of the normal at P is X¢ O X
ax sec q – by cosec q = a2 – b2 M
N
Now,
L = Ê a2 - b2 cos q, ˆ = (ae2cos q, 0) Y¢
ËÁ a 0˜¯
Thus the co-ordinates of M and N are
fi OL = ae2 cos q
Ê Ê b2 - a 2 ˆ ˆ
Thus, S¢L = ae + ae2 cos q = a3 (1 + e cos q), ÁË ËÁ b ˜¯ q¯˜
0, sin
SL = ae – ae2 cos = ae(1 – e cos q) a2 b2
Now, S ¢P = 1 + e cos q = S ¢L and Ê Ê - ˆ cos q, ˆ
SP 1 - e cos q SL ËÁ ËÁ a ¯˜ 0¯˜
fi PL bisects the –S¢PS internally. Ê b2 - a2 ˆ 2
ËÁ b b¯˜
Since PL ^ PT, therefore, PT will bisect –S¢PS exter- PM = a2cos2q + - sin 2q
nally.
44. Let the ellipse be x2 + y2 =1 and P(a cos q, b sin q) = a2cos2q + a4 sin 2q
a2 b2 b2
be any point on the ellipse. = a2 cos2q sin 2q
a2 b2
+
Ellipse 5.35
Ê a2 - b2 ˆ 2 Ê a tan q - b tan q ˆ
ËÁ a ¯˜ Á b 1+ a ˜
Also, PN = a - cos2q + b 2sin 2q \ tan j = ËÁ ˜¯
tan 2q
= b4cos2q + b 2sin 2q = Ê a2 - b2 ˆ tan q
a2 ÁË ab ˜¯ sec2q
= b2 cos2q + sin 2q = Ê a2 - b2 ˆ ¥ sin 2q
a2 b2 ÁË 2ab ˜¯
Thus, it is maximum when
Thus, PM : PN = a2 : b2 2q = 90° fi q = 45°.
Hence, the result.
46. Since the ordinate P meets the circle at Q, the co-ordi- Therefore, the maximum value is
nates of P and Q are (a cos q, b sin q) and (a cos q, a Ê a2 - b2 ˆ
sin q), respectively. ËÁ 2ab ¯˜
Y 48. The equation of any normal to the ellipse at P(q) is
Q ax secq – by cosec q = a2 – b2 …(i)
P
Let C be the centre of the ellipse.
X¢ ON X Then CM = Length of perpendicular from the centre C
R to the normal
= (a2 - b2)
Y¢
a2sec2q + b2cosec2q
The equation of the normal to the ellipse at P(q) is
= (a2 - b2)
ax sec q – by cosec q = a2 – b2 …(i) a2 + b2 + a2tan2q + b2cot2q
and the equation of normal to the auxiliary circle at Q(a < (a2 - b2) = a - b
cos q, a sin q) is a2 + b2 + 2ab
y = (tan q)x …(ii) 49. The equation of tangent at P is
x cos j + y sin j = 1
Solving, we get ab …(i)
…(ii)
cos q = x and sin q = y and the equation of normal at P is
(a + b) (a + b) ax sec j – by cosec j = a2 – b2
Squaring and adding, we get \ Q = (a sec f, 0)
x2 + y2 = (a + b)2
and R = Ê (a2 - b2 ) cos j ˆ
47. The equation of any normal at P(q) is ËÁ a , 0˜¯
x sec q – by cosec q = a2 – b2.
\ Slope of normal = m1 = a tan q
b
P
and slope of CP = m2 = b tan q. R aQ
a
P
Cf Therefore,
QR = a
N
fi a sec j - (a2 - b2 ) cos j = a
Let f be the angle between CP and the normal at P. a
fi a2 sin2 j + b2 cos2 j = a2 cos j
5.36 Coordinate Geometry Booster
P(h, k)
fi a2 sin2 j + a2(1 – e2) cos2 j = a2 cos j
fi a2(sin2 + cos2) – a2e2cos2 j = a2cos j
fi a2 – a2e2 cos2 = a2 cos j
fi c2 cos2 j + cos j – 1 = 0 QR
50. The equation of the normal to the ellipse x2 + y2 =1 90°
at P(x1, y1), Q(x2, y2) and R(x2, y3) are a2 b2 C
a2x - b2 y = a2 - b2 …(i) It subtends a right angle at the centre (0, 0) of the
x1 y1
ellipse
a2x - b2 y = a2 - b2
x2 y2 …(ii) x2 + y2 =1 …(ii)
a2 b2
and a2x - b2 y = a2 - b2
x3 y3 …(iii) From Eqs (i) and (ii), we get
Eliminating x and y from Eqs (i), (ii) and (iii), we get x2 + y2 = Ê xh + yk ˆ 2
a2 b2 ËÁ a2 b2 ¯˜
a2 - b2 (a2 - b2)
x1 y1 fi Ê1 - h2 ˆ x2 + Ê1 - k2 ˆ y2 - 2hxky = 0
a2 - b2 (a2 - b2) = 0 ÁË a2 a4 ˜¯ ËÁ b2 b4 ¯˜ a2b2
x2 y2
a2 - b2 (a2 - b2) Since these lines are at right angles, therefore sum of
x3 y3 the co-efficients of x2 and y2 is zero.
fi Ê1 - h2 ˆ + Ê1 - k2 ˆ = 0
ËÁ a2 a4 ˜¯ ÁË b2 b4 ¯˜
1 11 fi Ê h2 + k2 ˆ = 1 + 1
x1 y1 ËÁ a4 b4 ¯˜ a2 b2
fi 1 1 1=0
x2 y2 Hence, the locus of (h, k) is
1 11
x3 y3 Ê x2 + y2 ˆ = 1 + 1
ËÁ a4 b4 ˜¯ a2 b2
x1 y1 x1y1 52. Let the point from which the tangents are drawn be (h,
fi x2 y2 x2 y2 = 0 k).
So the equation of the chord of contact from the point
x3 y3 x3 y3
(h, k) to the given ellipse is
Also b sin a ab cos a sin a hx + ky =1 …(i)
a cos a b sin b ab cos b sin b = 0 a2 b2
a cos b b sin g ab cos g sin g
a cos g P(h, k)
cosec a sec a 1
fi cosec b sec b 1 = 0 QM R
cosec g sec g 1 c
C
sec a cosec a 1 It touches the circle x2 + y2 = c2.
fi sec b cosec b 1 = 0 Therefore, the length of the perpendicular from the
centre (0, 0) to the chord of contact (i) is equal to the
sec g cosec g 1 radius of the circle.
Hence the result.
51. Let the point from which tangents are drawn be (h, k). Thus, 0+ 0-1 = c
The equation of the chord of contact from the point (h, h2 k2
a4 b4
k) to the given ellipse is hx + ky = 1. +
a2 b2
Ellipse 5.37
h2 k2 1 From Eqs (i) and (ii), we get
a4 b4 c2
fi + = h k 1
cos q c sin +
c = q = h2 k2
a2 b2
Hence, the locus of (h, k) is
x2 + y2 = 1 hÊ h2 k2 ˆ
a4 b4 c2 c ËÁ a2 b2 ¯˜
fi cos q = +
53. Let (x1, y1) be the point of intersection of the perpendic- kÊ h2 k2 ˆ
c ÁË a2 b2 ¯˜
ular tangents, so that (x1, y1) lies on the director circle and sin q = +
x12 + y12 = a2 + b2 …(i)
The equation of the chord of contact from (x1, y1) is Squaring and adding, we get
xx1 yy1 ÊhÊ h2 k2 ˆ ˆ 2 Êk Ê h2 k2 ˆˆ 2
a2 b2 ÁË c ËÁ a2 b2 ˜¯ ˜¯ ÁË c ËÁ a2 b2 ˜¯ ˜¯
+ =1 …(ii) + + + =1
Let its mid-point be (h, k). k2 ˆ 2
b2 ¯˜
\ The equation of the chord bisected at (h, k) is fi Ê h2 + = Ê c2 ˆ
ÁË a2 ËÁ h2 + k 2 ¯˜
hx + ky = h2 + k2 …(iii)
a2 b2 a2 b2
Hence, the locus of (h, k) is
From Eqs (ii) and (iii), we get Ê x2 y2 ˆ 2 Ê c2 ˆ
ÁË a2 b2 ¯˜ ËÁ x2 + y2 ˜¯
x1 = y1 = 1 + =
h k +
Ê h2 k2 ˆ
ÁË a2 b2 ˜¯ 55. Let the co-ordinates of P be (h, k).
Now from Eqs. (i), we get Y
Ê h ˆ2 + Ê k ˆ2 = (a2 + b2 ) B P(h, k)
Á + ˜ Á +
ÁËÁ Ê h2 k2 ˆ ˜˜¯ ÁÁË Ê h2 k2 ˆ ˜ X¢ O AX
ËÁ a2 b2 ¯˜ ÁË a2 b2 ˜¯ ˜¯˜
Ê h2 k2 ˆ 2
ÁË a2 b2 ¯˜
fi (a2 + b2 ) + = (h2 + k2)
Y¢
Hence, the locus of (h, k) is The equation of its chord of contact with respect to the
ellipse
Ê x2 y2 ˆ 2
ÁË a2 b2 ¯˜
(a2 + b2 ) + = (x2 + y2) x2 y2
a2 b2
54. Let the point be (c cosq, c sinq). + =1 is
The equation of the tangent to the ellipse at (c cosq, c hx + ky =1
sinq) is a2 b2
x ◊ c cos q + y ◊ c cos q =1 …(i) It meets the axes in
a2 b2
Ê a2 ˆ Ê b2 ˆ
Let the co-ordinates of the mid-point be (h, k). A ËÁ h , 0˜¯ and B ËÁ 0, k ¯˜
P 1
2
Now, area of the triangle OAB = ◊ OA ◊ OB
O M(h, k) R = 1 ◊ a2 ◊ b2
2h k
= a2b2
2◊h◊k
\ The equation of the chord bisected at (h, k) is l
= constant = (say)
hx + ky = h2 + k2 …(ii)
a2 b2 a2 b2 2
Hence, the locus of (h, k) is x ◊ y = a2b2
l
which represents a hyperbola.
5.38 Coordinate Geometry Booster
56. Let the point P be (a cos q, b sin q). 58. Let the point be (h, k).
\ The equation of normal at P to the ellipse is
a ◊ x ◊ sec q – b ◊ y ◊ cosec q = a2 – b2 B
…(i)
C
90° M(h, k)
P A
M(h, k) T The equation of the chord bisected at (h, k) to the el-
N lipse is hx + ky = h2 + k2
a2 b2 a2 b2
Let its mid-point be (h, k). Ê Ê hx + ky ˆ ˆ
Á ËÁ a2 + b2 ¯˜ ˜
\ The equation of the chord bisected at (h, k) is fi Á h2 k2 ˆ ˜ = 1
Á Ê a2 b2 ¯˜ ˜
hx + ky = h2 + k2 …(ii) ÁË ÁË ˜¯
a2 b2 a2 b2
From Eqs (i) and (ii), we get Now we make it a homogenous equation of 2nd degree.
a sec q = b cosec q = a2 - b2 Ê Ê hx + ky ˆ ˆ2
(h/a2 ) (- k /b2 ) Ê h2 + b2 ¯˜ ˜
ÁË a2 + k2 ˆ Thus, Ê x2 + y2 ˆ = Á ÁË a2 k2 ˆ ˜
b2 ˜¯ ËÁ a2 b2 ˜¯ Á Ê h2 b2 ˜¯ ˜
Á ˜¯
ÁË ÁË a2
fi cos q = Ê h2 + k2 ˆ / Ê h ˆ
ËÁ a2 b2 ˜¯ ËÁ a3 (a 2- b2 ) ˜¯ k2 ˆ 2
Ê x2 y2 ˆ Ê h2 b2 ˜¯ Ê hx ky ˆ 2
Ê h2 k2 ˆ Ê - k ˆ fi ÁË a2 + b2 ˜¯ ËÁ a2 + = ËÁ a2 + b2 ˜¯
ÁË a2 b2 ˜¯ ÁË (a2 - ˜¯
and sin q = + / b3 b2 ) h2x2 k 2 y2
a4 b4
Squaring and adding, we get = + + 2hkxy
a2b2
Ê a6 + b6 ˆ Ê x2 + y2 ˆ 2 = (a2 - b2 )2 Since the chord subtends right angle at the centre, so
ÁË x2 y2 ˜¯ ËÁ a2 b2 ˜¯ co-efficient of x2 + co-efficient of y2 = 0
57. Let the point be (h, k). a 2 x2 + b 2 y2 = Ê x2 + y2 ˆ 2
a4 b4 ËÁ a2 b2 ˜¯
The equation of the chord bisected at (h, k) to the el-
lipse x2 + y2 =1 is fi Ê h2 + k2 ˆ2 Ê 1 + 1ˆ = Ê h2 + k2 ˆ
a2 b2 ËÁ a2 b2 ˜¯ ËÁ a2 b2 ¯˜ ÁË a4 b4 ¯˜
hx + ky = h2 + k2 …(i) Hence, the locus of (h, k) is
a2 b2 a2 b2
Ê x2 y2 ˆ 2 Ê x2 y2 ˆ
and the equation of the tangent to the auxiliary circle x2 ÁË a2 + b2 ˜¯ Ê 1 + 1ˆ = ÁË a4 + b4 ˜¯
+ y2 = a2 at (a cos qc a sin q) is ÁË a2 b2 ˜¯
x cos q + y sin q = a …(ii) 59. Let the point P be (a cos q, b sin q) and Q be
From Eqs (i) and (ii), we get Ê cos ËÊÁ p q¯ˆ˜ Ê p qˆ¯˜ ˆ
ËÁ 2 ËÁ 2 ¯˜
cos q = sin q = a a + , b sin + ,
h/a2 k /b2 +
Ê h2 k2 ˆ
ËÁ a2 b2 ¯˜ i.e. (– a sin qs b cos q).
Squaring and adding, we get Q q+ p
2
Ê 2 k2 ˆ Ê h2 k2 ˆ 2 P(q)
ËÁ 4 b4 ˜¯ ËÁ a2 b2 ˜¯
a2 h + = + M(h, k)
a
Hence the locus of (h, k) is
a2 Ê x2 + y2 ˆ = Ê x2 + y2 ˆ 2 Let M(h, k) be the mid-point of PQ.
ÁË a4 b4 ˜¯ ÁË a2 b2 ¯˜
Ellipse 5.39
Then, h = a (cos q - sin q ) Hence, the locus of (a, b) is
2
Ê x2 + y2 ˆ = Ê x2 + y2 ˆ 2
and k = b (cos q + sin q ) ËÁ a2 + b2 ˜¯ ËÁ a2 b2 ¯˜
2
61. Let the mid-point be (h, k).
fi 2h = (cos q - sin q ) and 2k = (cos q + sin q )
ab P
Squaring and adding, we get
Ê 4h2 + 4k 2 ˆ = 2 S(æ, 0)
ËÁ a2 b2 ¯˜ M(h, k)
fi Ê h2 + k2 ˆ = 1 Q
ÁË a2 b2 ¯˜ 2
Hence, the locus of (h, k) is The equation of the chord of the ellipse whose mid-
point (h, k) is
Ê x2 + y2 ˆ = 1 hx + ky = h2 + k2
ÁË a2 b2 ˜¯ 2 a2 b2 a2 b2
60. Equation of the chord of contact to the tangents at (h, k) which passes through the focus (ae, 0).
is Thus, hae h2 k2
a2 a2 b2
hx + ky =1 …(i) = +
a2 b2
he b2h2 + a2k2
P(h, k) fi a = a2b2
fi b2h2 + a2k2 = ab2he
Hence, the locus of (h, k) is
b2h2 + a2y2 = ab2xe
Q R 62. Let the point be (h, k).
M(a ◊ b)
A M(h, k) B
P(q)
The equation of the chord of the ellipse whose mid-
point (a, b) is
ax + by = a2 + b2 …(ii)
a2 b2 a2 b2
The equation of the chord of the ellipse whose middle
Since Eqs (i) and (ii) are the same, therefore point (h, k) is
h= k = Ê a2 1 b2 ˆ hx + ky = h2 + k2 …(i)
a b ÁË a2 + b2 ˜¯ a2 b2 a2 b2
and the equation of the tangent to the ellipse
fi h= a x2 + y2 =1 at (a cos q, b sin q) is
a2 b2 ˆ a2 b2
Ê a2 + b2 ˜¯
ÁË x cos q + y sin q = 1
ab
and k = b b2 ˆ Since both the equations are identical, so
Ê a2 b2 ˜¯
ËÁ a2 + cos q /a sin q /b 1
h/a2 k /b2 +
= = Ê h2 k2 ˆ
ÁË a2 b2 ¯˜
Also, (h, k) lies on the director circle of the given
ellipse x2 + y2 = a2 + b2. Squaring and adding, we get
Thus, h2 + k2 = a2 + b2 a 2h2 b 2k 2 Ê h2 k2 ˆ 2
a4 b4 ÁË a2 b2 ˜¯
Êa2 + b2ˆ Êa2 b2 ˆ 2 + = +
ËÁ a2 + b2 ¯˜ ÁË a2 b2 ¯˜
fi = +
5.40 Coordinate Geometry Booster
Hence the locus of (h, k) is 67. Let the pole be (h, k).
The equation of polar w.r.t. the ellipse c2x2 + d2y2 = 1 is
a 2 x2 b 2 y2 Ê x2 y2 ˆ 2
a4 + b4 = ËÁ a2 + b2 ¯˜ c2hx + d2ky = 1
x2 y2 fi d2ky = –c2 hx + 1
a2 b2
63. Let the ellipse be + =1 and its focus is (ae, 0). c2h 1
d 2k d 2k
fi y = - x +
The equation of the polar is x2 y2
a2 b2
xx1 + y ◊ y1 =1 which is a tangent to the ellipse + =1
a2 b2
So, c2 = a2m2 + b2
fi x (ae) + y (0) =1 1 Ê c2h ˆ 2
a2 b2 (d 2k)2 ËÁ d 2k ¯˜
fi = a 2 - + b2
fi x=a fi a2c4h2 + b2d1k2 = 1
e
which is the directrix. Hence, the locus of (h, k) is
a2c4x2 – b2d4y2 = 1
64. Let the pole be (x1, y1).
68. Let the pole be (h, k).
xx1 y ◊ y1
The equation of the polar is a2 + b2 =1 which is The equation of the polar is hx + ky =1
a2 b2
identical with lx + my + n = 0
0+ 0-1 = c
So, x1/a2 = y1/b2 = -1 It is given that h2 k2
l mn a4 b4
+
fi x1 = - a2l , y1 = - b2m h2 + k2 = 1
n n a4 b4 c2
fi
Hence, the pole is Ê - a2l , - b2m ˆ Hence, the locus of (h, k) is
ËÁ n n ˜¯
x2 y2 1
65. If (h, k) be the pole of a given line w.r.t. the ellipse, then a4 + b4 = c2
its equation is 69. Let the pole be (h, k).
hx + ky =1 …(i) The equation of the polar is hx + ky =1 …(i)
a2 b2 a2 b2 …(ii)
If tangents at its extremities meet at (a, b), then it is the and the equation of the normal to the ellipse is
chord of contact of (a, b) and hence its equation is ax sec j – by cosec j = a2 – b2
ax + by =1 …(ii) Comparing Eqs (i) and (ii), we get
a2 b2
Comparing Eqs (i) and (ii), we get h/a2 = - k /b2 = 1
a sec j cosec j -
h/a2 = k /b2 =1 b a2 b2
a /a2 b /b2
fi h cos j = - k sin j = 1
a3 b3cosec j -
fi h = a, k = b a2 b2
Thus, the pole (h, k) is the same as (a, b), i.e. the inter- fi cos j = a3 b2 ) , sin j = - b3 b2 )
section of tangents. h(a2 - k(a2 -
66. Let the pole be (h, k).
The equation of polar w.r.t. the ellipse x2 + 4y2 = 4 is Eliminating f, we get
hx + 4ky = 4 Ê a3 ˆ 2 Ê b3 ˆ 2
ËÁ 2- ˜¯ ÁË 2- ¯˜
which is identical with x + 4y = 4 h(a b2 ) + - k (a b2 ) = 1
So, h = 4k = 4
fi a6 + b6 = (a2 - b2 )2
144 h2 k2
fi h = k =1
Hence, the locus of (h, k) is a6 + b6 = (a2 - b2)2
11 x2 y2
Hence, the pole is (1, 1)
Ellipse 5.41
70. Let the pole be (h, k) Q(–a cos j, b sin j),
S(–a sin j, b cos j)
The equation of the polar w.r.t. the parabola is R(a sin j, –b cos j)
yk = 2a(x + h) and
73.
fi yk = 2ax + 2ah
fi y = 2a x + 2ah D P
which is a kk x2 + y2 =1 C
tangent to the ellipse a2 b2
Ê 2ah ˆ 2 Ê 2a ˆ 2
ËÁ k ˜¯ ÁË k ¯˜
So, = a 2 + b 2
fi 4a2h2 = 4a2a2 + k2b 2 Let CP and CD be two conjugate semi-diameters of an
Hence, the locus of pole (h, k) is ellipse axth22e+ecbyc22en=tr1icaanndglleetotfhDe eicscepnt+rijc angle of P is
f. Thus .
4a2x2 = y2b 2 + 4a2a2 2
71. Therefore the co-ordinates of P and D are (a cos j, b
B sin j) and
M(h, k) Ê a cos Ê p + j ˆ¯˜ , b sin Ê p + j ˆ ˆ
ËÁ ËÁ 2 ËÁ 2 ¯˜ ¯˜
A¢ A i.e. (–a sin j, b cos j)
Thus CP2 + CD2
y = mx + c
B¢ = (a2 cos2 j + b2 sin2 j) + (a2 sin2 j + b2 cos2 j)
= a2 + b2
Let (h, k) be the mid-point of the chord y = mx + c of the 74.
x2 y2 D P
a2 b2
ellipse + = 1.
Then T = S1 S¢(–a e, 0) C S(–a e, 0)
fi xh + yk = h2 + k2
a2 b2 a2 b2
fi k = - b2h Let CP and CD be the conjugate diameters of the
a2m ellipse.
b2x
Hence, the locus of the mid-point is y = - a2m Let P = (a cos j, b sin j). then the co-ordinates of D is
(–a sin j, b cos j).
72. Two diameters are said to be conjugate when each bi-
Thus,
sects all chords parallel to the other.
SP ◊ S¢P = (a – ae cos j)(a + ae cos j)
If y = m1x and y = m2x be two conjugate diameters of an = a2 – a2e2 cos2 j
= a2 – (a2 – b2) cos2 j
ellipse, then m1m2 = - b2 . = a2 sin2 + b2 cos2 j
a2 = CD2
SP
75 Let the eccentric angle of P is (f) and the eccentric
C angle of M is ÊÁË j + p ˜ˆ¯ .
2
QR
Then the co-ordinates of P and M are (a cos j, b sin j)
Let PQ and RS be two conjugate diameters. Then the and
co-ordinates of the four extremities of two conjugate
diameters are Ê a cos ÊËÁj + p ˆ , b sin ËÊÁ j + p ˆˆ
ËÁ 2 ¯˜ 2 ¯˜ ¯˜
P(a cos j, b sin j),
i.e. (a cos j, b sin j) and (–a sin j, b cos j)
5.42 Coordinate Geometry Booster
The equation of the tangent at P is …(i) fi m2 = - b3
x cos j + y sin j = 1 …(ii) a3
ab b3
y = - a3 x
and the equation of the tangent at M is Hence, the required diameters is
x sin j + y cos j = 1 fi a3y + b3x = 0
—a b
78. The given ellipses are
Squaring and adding Eqs (i) and (ii), we get
x2 y2
x2 + y2 = 2 a2 + b2 =1 …(i)
a2 b2
x2 y2 x2 y2
76. Let the equation of the ellipse be a2 + b2 = 1. and c2 + d2 =1 …(ii)
B The equation of lines passing through the point of in-
DP tersection of the ellipses are
M(h, k) Ê 1 1ˆ Ê 1 1ˆ
ÁË a2 c2 ˜¯ ÁË b2 d 2 ˜¯
x2 - + y 2 - = 0 …(iii)
A A
C
which represents a pair of lines through the origin.
QR If y = mx be one of the lines, then y = mx must satisfy
B¢
(iii), then
Let the eccentric angle of P is (f) and the eccentric Ê1 - 1ˆ + m2 Ê 1 - 1ˆ = 0
ËÁ a2 c2 ˜¯ ËÁ b2 d 2 ˜¯
angle of D is ÁËÊ j + p ¯ˆ˜ .
2 Ê1 1ˆ 1 1ˆ
fi ÁË b2 - d 2 ¯˜ m2 + Ê a2 - c2 ˜¯ = 0
Then the co-ordinates of P and D are (a cos j, b sin j) ÁË
and which is a quadratic in m. Let it has roots m1 and m2.
Ê a cos ÊÁËj + p ˆ˜¯ , b sin ÊËÁ j + p ˆˆ Ê1 - 1 ˆ
ËÁ 2 2 ˜¯ ¯˜ ËÁ c2 ¯˜
Ê a2
i.e. (a cos j, b sin j) and (–a sin j, b cos j) Then, m1m2 = - 1 1 ˆ
d2 ˜¯
Let M(h, k) be the mid-point of PD. ËÁ b2 -
Then
Ê1 1 ˆ
h = a cos j - a sin j b2 ÁË a2 - c2 ¯˜
2 a2 Ê1
fi - = ËÁ b2 1 ˆ
d2 ¯˜
a cos j + b sin j (2h)2 (2k )2 -
2 a2 b2
and k = fi +
= (cos j – sin j)2 + (cos j + sin j)2 fi a 2 Ê 1 - 1ˆ = - b2 Ê 1 - 1ˆ
ÁË a2 c2 ¯˜ ÁË b2 d 2 ˜¯
fi h2 + k2 = 1 fi Ê a2 ˆ = Ê b2 ˆ
a2 b2 2 ÁË1 - c2 ˜¯ ËÁ -1 + d2 ¯˜
x2 y2
Hence, the locus of (h, k) is a2 + b2 = 1 a2 b2
2 c2 d2
fi + = 2
77. The equation of the given diameter is
ax – by = 0 Hence, the result.
79. The co-ordinates of P and D are (a cos j, b sin j) and
fi y=ax …(i)
b (–a sin j, b cos j)
Let PM is the normal at P and DN is the normal at D.
Thus, m1 = a
b B
D
Let the diameter conjugate to (i) be y = m2x P
As we know that, two diameters y = m1x and y = m2x M HN
A¢ C
are conjugate, if m1 ◊ m2 = - b2 A
a2
b2
fi a ¥ m2 = - a2
b
B¢
Ellipse 5.43
The equations of the normals at P and D are Area of the parallelogram = MNM¢N¢
ax sec j – by cosec j = a2 – b2 = 4 (the area of the parallelogram CPMR)
and –ax cosec j – by sec j = a2 – b2 = 4 ¥ a2cos2j + b2sin2j ¥ ab
a2cos2j + b2sin2j
respectively. = 4ab
= constant
Since, H(a, b) is the point of intersection of the nor-
mals, so
aa sec j – bb cosec j – (a2 – b2) = 0 …(i) Hence, the result. ËÁÊ p + j¯˜ˆ
2
and 81. Let the eccentric angles at P and Q be j and
aa cosec j + bb sec j + (a2 – b2) = 0 respectively
…(ii)
Eliminating f from Eqs (i) and (ii), we get The equation of the tangents at P and Q
sec j = cosec j = (a2 - b2 ) are x cos j + y sin j = 1
(aa - bb) -(aa + bb ) (a2a 2 + b2b2 ) ab
fi cos j = (a2a 2 + b2b 2 ) and - x sin j + y cos j = 1
(a2 - b2 )(aa - bb) ab
respectively
(a2a2 + b2 b2 ) Squaring and adding, we get
(a2 - b2 )(aa + bb)
and sin j = - x2 + y2 = 2
a2 b2
Squaring and adding, we get
Hence, the result.
2(a2a2 + b2b2)3 = (a2 – b2)(a2a2 – b2b2)2
82. Thus, the equation of the ellipse is
Hence, the locus of H(a, b) is y2 + 2 x2 = 1
2(a2x2 + b2y2)3 = (a2 – b2)(a2x2 – b2y2)2. 3
80. fi x2 + y2 = 1
32
Y
RP Hence, the eccentricity is
X¢ X e= 1- b2 = 1- 2 = 1
C a2 3 3
QS 83. The given equation of an ellipse is
16x2 + 25y2 = 400
Y¢ fi x2 + y2 = 1
25 16
Let PCQ and RCS be two conjugate diameters of the
From the reflection property of an ellipse, the reflection
ellipse x2 + y2 = 1. ray passes through the focus.
a2 b2
Then the co-ordinates of P, Q, R, and S are Thus, the co-ordinates of the other focus = (3, 0).
P(a cos j, b sin j), Q(–a cos j, –b sin j), When y = 4, then x = 0.
So the point is (0, 4).
R(–a sin j, b cos j) and S(a sin j, –b cos j) respec-
tively. So the equation of the reflection ray is
The equations of tangents at P, R, Q and S are x + y =1
34
x cos j + y sin j = 1, fi 4x + 3y = 12
ab
84. The equation of the incident ray is
- x sin j + y cos j = 1, x–y+2=0
ab
fi x + y =1
- x cos j - y sin j = 1, -2 2
ab
The equation of the ellipse is
and x sin j - y cos j = 1 3x2 + 4y2 = 12
ab
fi x2 + y2 = 1
Thus, the tangents at P and Q are parallel. Also the tan- 43
gents at R and S are are parallel. Hence, the tangents at
P, R, Q, S form a parallelogram.
5.44 Coordinate Geometry Booster
Clearly, the co-ordinates of the foci are (–2, 0) and fi 1 = cos2q + sin 2q
(2, 0). d2 a2 b2
Since the incident ray x – y + 2 = 0 intersects the ellipse
at (0, 2), so, the equation of the reflection ray = the Now,
equation of the line joining (0, 2) and (2, 0)
fi x + y =1 4a2 Ê - b2 ˆ
ÁË1 d2 ˜¯
22
fi x+y=2 = 4a2 - 4a2b2
d2
LEVEL III = 4a2 - 4a 2b 2 Ê cos2q + sin2q ˆ
ËÁ a2 b2 ¯˜
1. Given ellipse is x2 + y2 =1 = 4a2 – 4b2 cos2q – 4a2sin2q
a2 b2 = 4a2(1 – sin2q) – 4b2cos2q
= 4a2cos2q – 4b2cos2q
So, F1 = (ac, 0) and F2 = (–ac, 0) = 4 cos2q(a2 – b2)
Let P be (a cos q, b sin q). = 4 cos2q(a2e2)
= (2ae cos q)2
Then = [(a + ae cos q) – (a – ae cos q)]2
= (PF1 – PF2)2
1 a cos q b sin q 1 3. Given ellipse is
2 0 1
ar(DPF1F2) = ae 0 1 x2 + y2 = 1
16 9
-ae
= 1 ¥ 2ae ¥ b sin q Ê ˆ
2 Foci = (± ae, 0) = ËÁ ± 4 ◊ 0¯˜
= abe ¥ sin q
Maximum value of A = abe
= ab 1 - b2 \ 7 , = (± 7, 0)
a2 4
= b a2 - b2 and the radius of the circle = 7 + 9 = 16 = 4
4. The equation of the tangent to the parabola y2 = 4x at
2. Given ellipse is (t2, 2t) is
x2 y2 yt = x + t2
a2 b2
+ =1 fi x – yt + t2 = 0 …(i)
Y The equation of the normal
4x2 + 5y2 = 20
M i.e. x2 + y2 = 1 at ( 5 cos j, 2 sin j)
P 54
X¢ is a2x - b2 y = a2 - b2
F2(a e, 0) F1(a e, 0) x1 y1
X
fi 5x - 4y = 5 — 4 = 1
5 cos j 2 sin j
Y¢ fi ( 5 sec j)x - (2 cosec j) y = 1 …(ii)
The equation of the tangent to the given ellipse at Since Eqs (i) and (ii) are the same line, so
P(a cos q, b sin q) is
x cos q + y sin q = 1 5 sec j = 2 cosec j = —1
ab 1 t t2
Now, OM = d fi sec j = - 1 , cosec j = - 1
5t 2 2t
fi 0+ 0-1 = d
cos2q sin 2q fi cos j = - t2 5, sin j = - 2t
a2 + b2
Ellipse 5.45
Thus, The equation of the chord of contact of the tangents
5t2 + 4t2 – 1 = 0 through A is
fi 5t4 + 5t2 – t2 – 1 = 0 hx + ky = 1 …(iv)
fi 5t2 (t2 + 1) – 1(t2 + 1) = 0 63
fi (5t2 – 1) (t2 + 1) = 0
fi (5t2 – 1) = 0 Since the eqs (ii) and (iv) are identical, so
fi t=± 1
hk
5
6 =3 =1
Also, tan j = ±2 cos q sin q
fi j = tan–1 (±2)
2
5. We have B = (0, b), F = (ae, 0) and F¢ = (–ae, 0)
Now, m(FB) = - b and m(BF ¢) = b fi h = 3 cos q, k = 3 sin q
ae ae
Now, squaring and adding, we get
h2 = k2 = 9
It is given that, Therefore, the locus of A is
m(FB) ¥ m(BF ¢) = –1
x2 + y2 = 9
fi - b ¥ b = -1 which is the director circle of x2 + y2 = 1
ae ae
63
fi b2 = a2e2
fi a2(1 – e2) = a2e2 Thus, the angle between the tangents at P and Q of the
fi (1 – e2) = e2 ellipse x2 + 2y2 = 6 is right angle.
fi 2e2 = 1
fi e= 1 7. Y
2 N
6. P
Y X¢ O X
A(h, k) L
Q Y¢
P Given ellipse is x2 + y2 =1
X¢ a2 b2
O X
Let P lies in the first quadrant, so
P = (a cos q, b sin q)
Y¢ The equation of the tangent at P is
x cos q + y sin q = 1
Given ellipse is ab
x2 + 4y2 = 4
Now, ON = -1
fi x2 + y2 = 1
41 …(i) cos2q + sin 2q
a2 b2
The equation of the tangent to the ellipse (i) is …(ii) = ab
x cos q + y sin q = 1 a2sin2q + b2cos2q
2
and the equation of the 2nd ellipse can be written as The equation of ON is
x2 + y2 = 1 (iii) x sin q - y cos q = 0
63 ba
and the equation of the normal at P is
Let the tangents at P and Q meet at A(h, k). ax sec q – by cosec q = a2 – b2
So PQ is the chord of contact.
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