MATHS COMBIND

Jpacademy Successive – Differential

y = ex sin x ⇒ yn = 2n/2 ex sin ⎛ x + nπ ⎞
⎝⎜ 4 ⎟⎠

DDD

Jpacademy CIRCLES

PREVIOUS EAMCET BITS

1. Th equation of the circles which passes through the origin and makes intercepts of length 4 and 8

on the x and y-axis respectively, are : [EAMCET 2009]

1) x2 + y2 ± 4x ± 8y = 0 2) x2 + y2 ± 2x ± 4 y = 0

3) x2 + y2 ± 8x ±16 y = 0 4) x2 + y2 ± x ± y = 0

Ans: 1

Sol. Equations of circles passing through (0, 0),(±4, 0),(0, ±8) are x2 + y2 ± 4x ± 8y = 0

2. The locus of centre of a circle which passes through the origin and cuts off a length of 4 units

from the line x = 3 is [EAMCET 2009]

1) y2 + 6x = 0 2) y2 + 6x = 13 3) y2 + 6x = 10 4) x2 + 6 y = 13

Ans: 2

Sol. OP = PM2 + 22

⇒ x2 + y2 = (x − 3)2 + 4

⇒ y2 + 6x = 13

3. The diameters of a circle are along 2x + y – 7 = 0 and x + 3y – 11 = 0. Then the equation of this

circle, which also passes through (5, 7) is [EAMCET 2009]

1) x2 + y2 − 4x − 6 y −16 = 0 2) x2 + y2 − 4x − 6 y − 20 = 0

3) x2 + y2 − 4x − 6 y −12 = 0 4) x2 + y2 + 4x + 6 y −12 = 0

Ans: 3
Sol. Centre of the circle = point of intersection of diameters = (2, 3) radius = distance between (5, 7)

and (2, 3) = 5

4. If the lines 2x – 3y = 5 and 3x – 4y = 7 are two diameters of a circle of radius 7, then the equation

of the circle is [EAMCET 2008]

1) x2 + y2 + 2x − 4y − 47 = 0 2) x2 + y2 = 49

3) x2 + y2 − 2x + 2y − 47 = 0 4) x2 + y2 = 17

Ans: 3

Sol. Centre of the circle = Point of intersection of {2x − 3y = 5,3x − 4y = 7} = (1, −1)

Equation of the circle = ( x −1)2 + ( y −1)2 = 72 ⇒ x2 + y2 − 2x + 2y − 47 = 0

5. The inverse of the point (1, 2) with respect to the circle x2 + y2 − 4x − 6y + 9 = 0 , is

[EAMCET 2008]

1) ⎜⎝⎛1, 1 ⎞ 2) (2, 1) 3) (0, 1) 4) (1, 0)
2 ⎟⎠

Ans: 3

Sol. The polar of (1, 2) is 1x + 2y − 2(x −1) − 3( y + 2) + 9 = 0

⇒ −x − y +1= 0 ⇒ x + y −1= 0

If (h, k) is the foot of the perpendicular then

h −1 = k − 2 = − (1+ 2 −1) = −1 ⇒ h = 0, k = 1⇒ Inverse point = (0, 1)
11 1+1

Jpacademy Circles

6. If θ is the angle between the tangents from (–1, 0) to the circle x2 + y2 − 5x + 4y − 2 = 0 , then θ =

[EAMCET 2008]

1) 2 tan −1 ⎛ 7 ⎞ 2) tan −1 ⎛ 7 ⎞ 3) 2 cot −1 ⎛ 7 ⎞ 4) cot −1 ⎛ 7 ⎞
⎜⎝ 4 ⎠⎟ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎠⎟

Ans: 1

Sol. Radius r = 25 + 4 + 2 = 7 S11 = 1+ 0 + 5 + 0 − 2 = 4
42

tan θ = r = 7 = 7 ⇒ θ = tan−1 7 ⇒ θ = 2 tan−1 7
2 S11 2 4 4 2 4 4

7. Th equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines x = 0

and y = 0 is [EAMCET 2007]

1) x2 + y2 − 6x + 6y + 9 = 0 2) x2 + y2 − 6x − 6y + 9 = 0

3) x2 + y2 + 6x − 6y + 9 = 0 4) x2 + y2 + 6x + 6y + 9 = 0

Ans: 1

Sol. Centre = (3, –3) radius = 3

Equation of the circle is x2 + y2 − 6x + 6y + 9 = 0

8. The inverse point of (1, 2) with respect to the circle x2 + y2 – 4x – 6y + 9 = 0 is [EAMCET 2007]

1) (0, 0) 2) (1, 0) 3) (0, 1) 4) (1, 1)

Ans: 3

Sol. Polar of (1, 2) w.r. to given circle is S1 = 0 ⇒ x + y −1 = 0 …….(1)

Foot of the perpendicular from (1, 2) to (1) is (0, 1)

∴ Inverse point is (0, 1)

9. Observe the following statements : [EAMCET 2006]

I) The circle x2 + y2 − 6x − 4y − 7 = 0 touches y-axis

II) The circle x2 + y2 + 6x + 4y − 7 = 0 touches x-axis.

Which of the following is a correct statement?

1) Both I and II are true 2) Neither I nor II is true

3) I is true, II is false 4) I is false, II is true

Ans: 2

Sol. For circle (1) f 2 − c = 4 + 7 = 11 > 0 ; For circle (2) g2 − c = 4 + 7 = 11 > 0

Both circles intersect X and Y axis
10. The length of the tangent drawn to the circle x2 + y2 − 2x + 4y −11 = 0 from the point (1, 3) is

[EAMCET 2006]
1) 1 2) 2 3) 3 4) 4
Ans: 3

Sol. Length of tangent = S11

= 1+ 9 − 2 +12 −11 = 3

11. If x – y + 1 = 0 meets the circle x2 + y2 + y – 1 = 0 at A and B, then the equation of the circle

with AB as diameter is [EAMCET 2005]

( )1) 2 x2 + y2 + 3x − y +1 = 0 ( )2) 2 x2 + y2 + 3x − y + 2 = 0

2

Jpacademy Circles

( )3) 2 x2 + y2 + 3x − y + 3 = 0 4) x2 + y2 + 3x − y +1 = 0

Ans: 1
Sol. S + λ L = 0

x2 + y2 + y −1+ λ (x − y +1)

Centre ⎛ −λ , ⎛ λ −1⎞⎞ lies on x – y + 1 = 0
⎝⎜ 2 ⎝⎜ 2 ⎠⎟ ⎟⎠

( )∴ λ = 3 , Required equation of circle is 2 x2 + y2 + 3x − y +1 = 0
2

12. If y = 3x is a tangent to a circle with centre (1, 1), then the other tangent drawn through (0, 0) to

the circle is : [EAMCET 2005]

1) 3y = x 2) y = –3x 3) y = 2x 4) y = – 2x

Ans: 1

Sol. By verification r = d is true for 3y = x.
13. If P1, P2, P3 are the perimeters of the three circles x2 + y2 + 8x − 6y = 0 .

4x2 + 4y2 − 4x −12y −186 = 0 and x2 + y2 − 6x + 6y − 9 = 0 respectively. Then [EAMCET 2004]

1) P1 < P2 < P3 2) P1 < P3 < P2 3) P3 < P2 < P1 4) P2 < P3 < P1
Ans: 2

Sol. Perimeter of the circle = 2πr

r1 = 5, r2 = 7, r3 = 3 3 ⇒ P1 < P3 < P2

14. If the line 3x –2y + 6 = 0 meets X-axis and Y-axis respectively at A and B, then the equation of

the circle with radius AB and centre at A is [EAMCET 2004]

1) x2 + y2 + 4x + 9 = 0 2) x2 + y2 + 4x − 9 = 0

3) x2 + y2 + 4x + 4 = 0 4) x2 + y2 + 4x − 4 = 0

Ans: 2

Sol. A (−2, 0), B = (0,3)

AB = 3

∴ Equation of the circle is ( x + 2)2 + y2 = 13

15 .If (1, a), (b, 2) are conjugate points with respect to the circle x2 + y2 = 25, then 4a + 2b =
[EAMCET 2004]

1) 25 2) 50 3) 100 4) 150

Ans: 2

Sol. S12 = 0 ⇒ b + 2a = 25
⇒ 4a + 2b = 50

16. If P is a point such that the ratio of the squares of the lengths of the tangents from P to the circles
x2 + y2 + 2x − 4y − 20 = 0 and x2 + y2 − 4x + 2y − 44 = 0 is 2 : 3, then the locus of P is a circle

with centre 2) (−7,8) 3) (7,8) [EAMCET 2003]

1) (7, −8) 4) (−7, −8)

Ans: 2
Sol. S11 : S1′1 = 2 : 3 = 3S = 2S′

⇒ x2 + y2 +14x −16y + 28 = 0

3

Jpacademy Circles

∴ Centre = (–7, 8)

17. If 5x – 2y + 10 = 0 and 12y – 5x + 16 = 0 are two tangents to a circle, then the radius of the circle

is [EAMCET 2003]

1) 1 2) 2 3) 4 4) 6

Ans: 1

Sol. Radius = 1/2 (distance between the given lines)

= 1 ⎛ 10 +16 ⎞ = 1
2 ⎝⎜ 13 ⎠⎟

18. The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is

[EAMCET 2002]

1) ( x − 5)2 + ( y + 5)2 = 25 2) ( x + 5)2 + ( y + 5)2 = 25

3) (x + 4)2 + ( y + 4)2 = 25 4) ( x + 6)2 + ( y + 6)2 = 25

Ans: 2
Sol. Centre (–a, –a) radius = a= 5

∴ ( x + 5)2 + ( y + 5)2 = 25

19. The radius of the larger circle lying in the first quadrant and touching the lines 4x + 3y – 12 = 0

and the coordinate axes is [EAMCET 2002]

1) 5 2) 6 3) 7 4) 8

Ans: 2
Sol. Centre (a, a), radius = a

For distance from centre (a, 0); To line 4x + 3y – 12 = 0

4a + 3a −12 = a ⇒ a = 1to 6
5

∴ Large radius = 6

20. The four distinct points (0, 0), (2, 0), (0, –2) and (k, -2) are concylic if k = [EAMCET 2002]

1) 2 2) – 2 3) 0 4) 1

Ans: 1

Sol. Equation of circle (0, 0), (a, 0), (0, b) is x2 + y2 − 2 + 2y = 0

Substituting (k, -2) in equation ⇒ k = 2

21. A line is at a constant distance C from the origin and meets the coordinate axes in A and B. The

locus of the centre of the circle passing through O, A, B is [EAMCET 2002]

1) x−2 + y−2 = c−2 2) x−2 + y−2 = 2c−2 3) x−2 + y−2 = 3c−2 4) x−2 + y−2 = 4c−2

Ans: 4

Sol. Let the line be x cos α + y sin α = c ∴ A ⎛ c α , 0 ⎞ ; B ⎜⎛⎝ 0, c α ⎞
⎜⎝ cos ⎟⎠ sin ⎟⎠

Midpoint of AB is centre of circle passing through O, A, B i.e ⎛ 2 c α , 2 c α ⎞
⎝⎜ cos sin ⎠⎟

∴ Locus of the centre is x−2 + y−2 = 4c−2

22. The equation of the normal to the circle x2 + y2 + 6x + 4y – 3 = 0 at (1, –2) is [EAMCET 2001]

1) y + 1 = 0 2) y + 2 = 0 3) y + 3 = 0 4) y – 2 = 0

Ans: 2

Sol. Centre of the circle = (–3, –2) ;

4

Jpacademy Circles

Normal passes through centre

∴ The normal at (1, –2) is y+ 2 = 0
23. If the polar of a point on the circle x2 + y2 = p2 with respect to the circle x2 + y2 = q2 touches

the circle x2 + y2 = r2 , then p, q, r are in ………progression [EAMCET 2001]

1) Arithmetic 2) Geometric 3) Harmonic 4) Arithmetico Geometric

Ans: 2
Sol. Let (x1, y1) is a point on x2 + y2 = p2 ⇒ x12 + y12 = P2

The Polar of ( x1, y1 ) w.r.t x2 + y2 = q2 is xx1 + yy1 = q2........(1)

(1) touches x2 + y2 = r2

⇒ q2 = r ⇒ q2 = pr
x12 + y12

24. The centre of the circle touching the y-axis at (0, 3) and making an intercept of 2 units on the

positive x-axis is [EAMCET 2000]

(1) 10, 3) 2) ( )3,10 3) ( )10,3 (4) 3, 10 )

Ans: 3
Sol. The circle touches y-axis at (0, 3)

∴ f = –3; c = 9

The length of intercept made by the circle on x-axis = 2 g2 − c = 2 ⇒ g = ± 10

( )∴ centre = 10,3

(∵ lies in the first Quadrant)
25. The slope m of a tangent through the point (7, 1) to the circle x2 +y2 = 25 satisfies the equation

[EAMCET 2000]

1) 12m2 + 7m −12 = 0 2)16m2 − 24m + 9 = 0

3) 12m2 − 7m −12 = 0 4) 9m2 + 24m +16 = 0

Ans: 3

Sol. The equation of the tangents is ( y −1) = m ( x − 7)

⇒ mx − y + (1− 7m) = 0.....(1)

(1) touches the circles x2 + y2 = 25

1− 7m = 5 ⇒ 12m2 − 7m −12 = 0
m2 +1

26. The number of common tangent that can be drawn to the circles x2 + y2 = 1 and

x2 + y2 − 2x − 6y + 6 = 0 is [EAMCET 2000]

1) 1 2) 2 3) 3 4) 4
Ans: 4

Sol. C1′ (0, 0) C2 (1,3) , r1 = 1; r2 = 2

C1C2 = 10 > r1 + r2

@@@@

5

Jpacademy ELLIPSE

PREVIOUS EAMCET BITS

1. If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8, then the

eccentricity is [EAMCET 2009]

1) 1 2) 1 3) 3 4) 4
5 2 5 5
2b = 8 ⇒ b = 4
Ans: 3
Sol: 2ae = 6 ⇒ ae = 3

( )∴ b2 = a2 1− e2

⇒ 16 = a2 − 9 ⇒ a = 5

∴ e=3
5

2. For an ellipse with eccentricity ½ the centre is at the origin. If one directrix is x = 4, then the

equation of the ellipse is [EAMCET2008]

1) 3x2 + 4y2 = 1 2) 3x2 + 4y2 = 12 3) 4x2 + 3y2 = 1 4) 4x2 + 3y2 = 12

Ans: 2

Sol. a = 4 ⇒ a = 4 ⇒ 2a = 4
e 1/ 2

⇒a=2 ( )b2 = a2 1− e2 = 4 ⎝⎛⎜1 − 1 ⎞ = 3
4 ⎟⎠

Equation of the ellipse is x2 + y2 = 1 ⇒ 3x2 + 4y2 = 12
43

3. The value of k if (1, 2), (k, –1) are conjugate points with respect to the ellipse 2x2 + 3y2 = 6 is

[EAMCET 2007]
1) 2 2) 4 3) 6 4) 8

Ans: 3

Sol. S12 = 0 ⇒ k = 6 [EAMCET 2006]
4. Equation of the latus recta of the ellipse 9x2 + 4y2 −18x − 8y − 23 = 0 are

1) y = ± 5 2) x = ± 5 3) y = 1± 5 4) x = −1± 5

Ans: 3

(x −1)2 ( y −1)2

Sol. Equation can be written as + = 1
49

Equation of latus recta of (x − α)2 + (y −β)2 =1 are

b2 a2

y = b ± ae y = 1± 5

5. The sides of the rectangle of greatest area that can be inscribed in the ellipse x2 + 4y2 = 64 are

( ) ( )1) 6 2,4 2 ( )3) 8 2,8 2 [EAMCET 2006]
2) 8 2, 4 2
( )4) 16 2, 4 2

1

Jpacademy Ellipse

Ans: 2

Sol. The sides of a rectangle of greatest area that can be inscribed in a ellipse x2 + y2 = 1 are given
a2 b2

equation x2 + y2 = 1
64 16

1=a 2 breadth = b 2

=8 2 42

6. The eccentricity of the conic 36x2 +144y2 − 36x − 96y −119 = 0 is [EAMCET 2004]

1) 3 2) 1 3) 3 4) 1
2 2 4 3

Ans: 1

Sol. e = 144 − 36 = 3 [EAMCET 2003]
144 2

7. The eccentricity of the ellipse 9x2 + 5y2 −18x − 20y −16 = 0 is

1) 1 2) 2 3) 3 4) 2
23 2

Ans: 2
Sol. 9x2 + 5y2 −18x − 20y −16 = 0

( x −1)2 ( y − 2)2
⇒+ =1
59

e= b2 − a2 = 2
b2 3

8. The pole of the straight line x + 4y = 4 with respect to the ellipse x2 + 4 y2 = 4is [EAMCET 2002]

1) (1, 4) 2) (4, 1) 3) (4, 4) 4) (1, 1)

Ans: 4

Sol. Pole = ⎛ −a 2A , −b2m ⎞
⎜ n n ⎟
⎝ ⎠

Ax + my + n = x + 4y − 4 = 0

x2 + y2 =1 [EAMCET 2001]
41
Pole = (1, 1)
9. The eccentricity of the ellipse x2 + y2 = 1 is

9 16

1) 7 2) 5 3) 7 4) 7
16 4 4 2

Ans: 3

Sol. e= b2 − a2 = 7
b2 4

10. The eccentricity of the ellipse 5x2 + 9y2 = 1 is [EAMCET 2000]

1) 2/3 2) 3/4 3) 4/5 4) 1/2

Ans: 1

3

Jpacademy Ellipse

Sol. e = 9 − 5 = 2
93

11. The product of the perpendicular from the foci on any tangent to the ellipse x2 + y2 = 1 is
a2 b2

[EAMCET 2000]

1) a 2) a2b2 3) b2 4) a2 + b2

Ans: 3
Sol. The product of ⊥ ers from foci to any tangent = b2

777

3

Jpacademy HYPERBOLA

PREVIOUS EAMCET BITS

1. If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points ( xi , yi ) , for i =1, 2, 3 and

4, then y1 + y2 + y3 + y4 = [EAMCET 2009]

1) 0 2) c 3) a 4) c4

Ans: 1

Sol. x = c2 ⇒ c4 + y2 = a2
y y2

⇒ y4 − a2y2 + c4 = 0

⇒ y1 + y2 + y3 + y4 = 0 [EAMCET 2009]
2. The midpoint of the chord 4x – 3y = 5 of the hyperbola 2x2 – y3 = 12 is

1) ⎛ 0, − 5⎞ 2) (2, 1) 3) ⎛ 5 , 0 ⎞ 4) ⎛ 11 , 2 ⎞
⎜⎝ 3 ⎠⎟ ⎝⎜ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠

Ans: 2

Sol. Write S1 = S11 and Compare [EAMCET 2008]
3. The distance between the foci of the hyperbola x2 − 3y2 − 4x − 6y −11 = 0

1) 4 2) 6 3) 8 4) 10
Ans: 3

Sol. x2 − 3y2 − 4x − 6y −11 = 0

( ) ( )⇒ x3 − 4x + 4 − 3 y2 + 2y +1 −12 ⇒ (x − 2)2 − 3( y +1)2 = 12

⇒ ( x − 2)2 − ( y +1)2 = 1 ⇒ a2 = 12, b2 = 4 ⇒ e = a2 + b2 = 12 + 4 = 4 = 1
12 4 a 44

( ) ( )b2 = a2 e1 −1 ⇒ 4 = 12 e2 −1 ⇒ e2 −1 = 1 ⇒ e2 = 4 ⇒ e = 2
33 3

( )Distance between the foci = 2ae = 2 12 2 / 3 = 8

4. If the line Ax + my = 1 is a normal to the hyperbola x2 − y2 = 1, then a2 − b2 = [EAMCET 2007]
a2 b2 A2 m2

1) a2 − b2 2) a2 + b2 ( )3) a2 + b2 2 ( )4) a2 − b2 2

Ans: 3

a2 + b2 2
n2
( )Sol.
a2 − b2 =
A2 m2

1

Jpacademy Hyperbola

( )⇒a2 − b2 2
A2 m2 = a2 + b2

5. If the eccentricity of a hyperbola is 3 , then the eccentricity of its conjugate hyperbola is
[EAMCET 2006]

1) 2 2) 3 3) 3 4) 2 3
2

Ans: 3

Sol. 1 + 1 =1
e2
(e′)2

1+ 1 =1⇒ 1 =1− 1 = 2
3 3 3
(e′)2 ( e′)2

e1 = 3
2

6. The product of the lengths of perpendicular drawn from any point on the hyperbola

x2 − 2y2 − 2 = 0 to its asymptotes is [EAMCET 2003]

1) 1 2) 2 3) 3 4) 2
23 2

Ans: 2

Sol. x2 −y2 = 1
21

Product of the length of perpendiculars from any point on the hyperbola to its asymptotes

a2b2 = 2
a2 + b2 3

7. If e and e′ are the eccentricities of the ellipse 5x2 + 9y2 = 45 and the hyperbola 5x2 − 4y2 = 45

respectively, then ee′ = [EAMCET 20002]

1) 9 2) 5 3) 4 4) 1

Ans: 4

Sol. Ellipse x2 + y2 = 1
95

⇒e= a2 − b2 = 9−5 = 2
a2 93

Hyperbola x2 − y2 = 1
9 45 / 4

⇒ e1 = a2 + b2 = 9 + 45 = 3
a2 4

92

3

Jpacademy Hyperbola

e.e1 = 2 . 3 = 1
32

8. [EAMCET 2002]
[EAMCET 2001]
1) 2) 3) 4)

Ans:

Sol.

9. The equation 16x2 + y2 + 8xy − 74x − 78y + 212 = 0 represents

1) A circle 2) A parabola 3) An ellipse 4) A hyperbola

Ans: 2

Sol. h2 − ab = 16 −16 = 0

10. The curve represented by x = 2(cost + sint) and y = 5 (cost – sint) is [EAMCET 2000]

1) a circle 2) a parabola 3) an ellipse 4) a hyperbola

Ans: 3

Sol. x = (cos t + sin t) = y = cos t = sin t

25

x2 + y2 = 2
4 25

VVV

3

Jpacademy PARABOLA

PREVIOUS EAMCET BITS

1. The number of normals drawn to the parobola y2 = 4x from the point (1, 0) is [EAMCET 2009]

1) 0 2) 1 3) 2 4) 3

Ans: 2

Sol. Given point is the focus, then no. of normals =1

2. If 2x + 3y + 12 = 0 and x – y + 4λ = 0 are conjugate with respect to the parabola y2 = 8x, then λ =
[EAMCET 2008]

1) 2 2) – 2 3) 3 4) – 3

Ans: 4

Sol. 2x + 3y + 12 = 0, x – y + 4λ = 0 are conjugate w.r.t the parabola y2 = 8x, then

⇒ 2(4λ) +1(12) = 2(2)(3)(−1) [ ]∵ 1n2 + 2n1 = 2am1m2

⇒ 8λ +12 = −12 ⇒ 8λ = −24

⇒ λ = −3
3. For the parabola y2 + 6y – 2x + 5 = 0 (I) The vertex is (–2, –3) (II) The directrix is y + 3 = 0

Which of the following is correct? [EAMCET 2007]

1) Both I and II are true 2) I is true, II is false

3) I is false, II is true 4) Both I and II are false

Ans: 2

Sol. ( y + 3)2 = 2(x + 2)

Vertex = (–2, –3); directrix = x = –3/2

4. If the lines 2x + 3y + 12 = 0 and x – y + 4k = 0 are conjugate with respect to the parabola y2 =8x,

then the value of k is [EAMCET 2006]

1) – 3 2) 3 3) 2 4) – 2

Ans: 1

Sol. Condition for lines to be conjugate is 1n2 + 2n1 = 2am1m2

a=2

2(4k) + 12 = 4(3) (–1)

k=-3

5. The parabola with directrix x + 2y – 1 = 0 and focus (1, 0) is [EAMCET 2005]

1) 4x2 − 4xy + y2 − 8x + 4y + 4 = 0 2) 4x2 + 4xy + y2 − 8x + 4y + 4 = 0

3) 4x2 + 4xy + y2 − 8x − 4y + 4 = 0 4) 4x2 − 4xy + y2 − 8x − 4y + 4 = 0

Ans: 1
Sol. SP2 = PM2

( x −1)2 + y2 = x + 2y −1 2

5

⇒ 4x2 − 4xy + y2 − 8x + 4y + 4 = 0

Jpacademy Parabola

6. The line, among the following, that touches the parabola y2 = 4ax is [EAMCET 2005]

1) x + my + am3 = 0 2) x − my + am2 = 0 3) x + my − am2 = 0 4) y + mx + am2 = 0

Ans: 2

Sol. Equation of tangent to the parabola y2 = 4ax is y = mx + a or y = 1 x + am
mm

7. The equation of the parabola with focus (0, 0) and directrix x + y = 4 is [EAMCET 2003]

1) x2 + y2 − 2xy + 8x + 8y −16 = 0 2) x2 + y2 − 2xy + 8x + 8y = 0

3) x2 + y2 + 8x + 8y −16 = 0 4) x2 − y2 + 8x + 8y −16 = 0

Ans: 1
Sol. SP = PM ⇒ SP2 = PM2

x2 + y2 = ⎛ x + y− 4 ⎞2 ⇒ x2 − 2xy + y2 + 8x + 8y − 16 = 0
⎝⎜ 2 ⎠⎟

8. A variable circle passes through the fixed point (2, 0) and touches the y-axis. Then locus of its

centre is [EAMCET 2002]

1) A parabola 2) A circle 3) An ellipse 4) A hyperbola

Ans: 1

Sol. The distance between the centre and pt(2, 0) is equal to the for distance from centre to y-axis.

∴ Locus of centre is parabola.

9. The equation of the parabola with the focus (3, 0) and the directrix x + 3 = 0 is [EAMCET 2002]

1) y2 = 3x 2) y2 = 6x 3) y2 = 12x 4) y2 = 2x

Ans: 3

Sol. Focus (a, 0) = (3, 0) a = 3

Directrix x + a = 0 ⇒ x + 3 = 0

∴ Equation of parabola is y2 = 4ax = 12x

10. Locus of the poles of focal chords of a parabola is………. of the parabola [EAMCET 2002]

1) the axis 2) a focal chord 3) the directrix 4) the tangent at the vertex

Ans: 3

Sol. Locus of poles of focal chords of Parabola is its directrix.

11. The length of the latus rectum of the parabola y2 + 8x − 2y +17 = 0 is [EAMCET 2001]

1) 2 2) 4 3) 8 4) 16
Ans: 3

Sol. Length of the L.R = 8 = 8
1

⎛ Coefficient of ' x ' ⎞
⎜ ⎟
⎝ Coefficient of ' y2 ' ⎠

12. If the normal to the parabola y2 = 4x at P(1, 2) meets the parabola again in Q, then Q =
[EAMCET 2001]

1) (–6, 9) 2) (9, –6) 3) (–9, –6) 4) (–6, –9)

Ans: 2

2

Jpacademy Parabola

( )Sol. Normal at at12, 2at1 meet the parabola y2 = 4ax

( )At at 2 , then t2 = −t1 − 2
2 2at 2 t1

Let 2at1 = 2t1 = 2(∵ a = 1) ⇒ t1 = 1

t2 = −1 − 2 = −3
1

( )∴ at 2 , = (9, −6)
2 2at 2

13. A variable circle passes through the fixed point (2, 0) and touches the y-axis. Then the locus of its

centre is [EAMCET 2000]

1) A parabola 2) A circle 3) An ellipse 4) A hyperbola

Ans: 1

Sol. Let the centre of the circle be (x1, y1) equation of the circle is S = x2 + y2 − 2x1x − 2y1y + c = 0
S = 0− , touches y-axis

∴ C = y12

∴ S = x2 + y2 − 2x1x − 2y1y + y12 = 0 it passes through (2, 0)4 – 4x1 + y12 = 0

∴ Locus of (x1, y1) is y2 = 4(x – 1) (parabola)

14. The vertex of the parabola x2 + 8x +12y + 4 = 0 [EAMCET 2000]

1) (–4, 1) 2) (4, –1) 3) (–4, –1) 4) (4, 1)

Ans: 1

Sol. x2 + 8x +12y + 4 = 0

⇒ (x + 4) 2 = −12( y −1)

∴ Vertex = (–4, 1) [EAMCET 200]
15. The line 4x + 6y + 9 = 0 touches the parabola y2 = 4x at the point.

1) ⎛ −3, 9 ⎞ 2) ⎛ 3, − 9 ⎞ 3) ⎛ 9 , −3⎟⎠⎞ 4) − ⎛ 9 , −3 ⎞
⎝⎜ 4 ⎠⎟ ⎝⎜ 4 ⎟⎠ ⎜⎝ 4 ⎝⎜ 4 ⎟⎠

Ans: 3

Sol. y = ⎛ −2 ⎞ x − 3 …………….(1)
⎝⎜ 3 ⎟⎠ 2

1 tangent to y2 = 4x

∴ Point of contact = ⎛ a ; 2a ⎞ = ⎛ 9 − 3 ⎞
⎝⎜ m2 m ⎠⎟ ⎜⎝ 4 ⎠⎟

DDD

3

Jpacademy POLAR COORDINATES

PREVIOUS EAMCET BITS

1. The eccentricity of the conic 5 = 2 + 3cos θ + 4sin θ is [EAMCET 2009]
r

1) 1 2) 1 3) 3 4) 5
2 22

Ans: 4

Sol. 5 = 1+ 5 cos (θ + α) ⇒ e = 5
2r 2 2

( )2. The radius of the circle with the polar equation r2 − 8r 3 cos θ + sin θ +15 = 0 is [EAMCET 2008]

1) 8 2) 7 3) 6 4) 5

Ans: 2

( )Sol. Given equation r2 − 8r 3 cos θ + sin θ +15 = 0

⇒ r 2 − 2r ( 8) ⎛ 3 cos θ + 1 sin θ ⎞ = −15
⎜⎜⎝ 2 2 ⎠⎟⎟

∴ a2 − c2 = −15 ⇒ a2 − 64 = −15 ⇒ a2 = 49 ⇒ a = 7

3. The area (in square units) of the triangle formed by the points with polar coordinates (1, 0),

(2, π/3) and (3, 2π/3) [EAMCET 2007]

1) 11 3 2) 5 3 3) 5 4) 11
4 4 4 4

Ans: 2

∑Sol. 1 r1r2 sin (θ1 − θ2 ) = 53
2 4

4. The polar equation of the circle with centre ⎛ 2, π⎞ and radius 3 units is [EAMCET 2006]
⎜⎝ 2 ⎟⎠

1) r2 + 4r cos θ = 5 2) r2 + 4r sin θ = 5 3) r2 − 4r sin θ = 5 4) r2 − 4rco s θ = 5

Ans: 3

1

Jpacademy
Polar Coordinates

Sol. Polar equation of circle with centre( c, α) and r is r2 + c2 − 2rc cos (θ − α) = a2

c ⎛ 2, π ⎞ , a = 3
⎜⎝ 2 ⎟⎠

r 2 + 4 − 4r cos ⎛ θ − π ⎞ = 9
⎜⎝ 2 ⎟⎠

r2 − 4r sin θ = 5

5. The cartesian form of the polar equation θ = tan−1 2 is [EAMCET 2005]

1) x = 2y 2) y = 2x 3) x = 4y 4) y = 4x

Ans: 2

Sol. θ = tan−1 2 ⇒ tan θ = 2 ⇒ y = 2 ⇒ y = 2x
x

6. Which of the following equations gives a circle? [EAMCET 2005]

1) r = 2sinθ 2) r2 cos 2θ = 1 ( )3) r (4 cos θ + 5sin θ) = 3 4) 5 = r 1+ 2 cos θ

Ans: 1
Sol. r2 = 2r sin θ

x2 + y2 = 2y which is a circle equation.

7. The polar equation cos θ + 7 sin θ = 1 represents a [EAMCET 2004]
r

1) Circle 2) Parabola 3) Straight line 4) Hyperbola

Ans: 3

Sol. ----

8. The centre of the circle r2 − 4r (cos θ + sin θ) − 4 = 0 in cartesian coordinates is [EAMCET 2004]

1) (1, 1) 2) (–1, –1) 3) (2, 2) 4) (–2, –2)

Ans: 3

Sol. Cartesian equation of the circle is x2 + y2 − 4x − 4y − 4 = 0

∴ Centre = (2, 2)

Jpacademy Polar Coordinates
9. The radius of the circle r = 3 sin θ + cos θ is [EAMCET 2004]
4) 4
1) 1 2) 2 3) 3

Ans: 1

Sol. radius = 3 + 1 = 1
44

10. The line passing through ⎛ −1, π ⎞ and perpendicular to 3 sin θ + 2 cos θ = 4 is [EAMCET 2003]
⎝⎜ 2 ⎠⎟ r

1) 2 = 3r cos θ − 2r sin θ 2) 5 = −2 3r cos θ + 4r sin θ

3) 2 = 3r cos θ + 2r sin θ 4) 5 = 2 3r sin θ + 4rcosθ

Ans: 1

Sol. equation of the line passing through ⎛ −1, π ⎞ and ⊥ er to 3 sin θ + 2 cos θ = 4 is
⎜⎝ 2 ⎟⎠ r

3 cos θ − 2sin θ = K
r

It passes through ⎛ −1, π ⎞ ⇒ 0 − 2 = k ⇒ k = 2
⎝⎜ 2 ⎟⎠ −1

∴ 3 cos θ − 2sin θ = 2
r

11. The equation 1 = 1 + 3 cos θ represents [EAMCET 2002]
r 88

1) A parabola 2) An ellipse 3) A hyperbola 4) A rectangular hyperbola

Ans: 3
Sol. A = 1+ e cos θ ⇒ 8 = 1+ 3cos θ

rr

e = 3 > 1 represents Hyperbola.
12. The equation of curve in polar coordinates is A = 2sin2 θ . Then it represents: [EAMCET 2001]

rr

1) A straight line 2) A circle 3) A parabola 4) An ellipse