﻿ IIT - JEE Main Important Questions of Conic Section | Zigya

## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Conic Section

#### Multiple Choice Questions

1.

If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2+ 4ax = and  x2+ 4ay = , then

• d2 + (2b+3c)2 = 0

• d2 +(3d+2c2) = 0

• d2 + (2b-3c)2 = 0

• d2 + (3b-2c)2 = 0

A.

d2 + (2b+3c)2 = 0

The equation of parabolas are y2 = 4ax and x2 = 4ay
On solving these we get x = 0 and x = 4a Also y = 0 and y = 4a
∴ The point of intersection of parabolas are A(0, 0) and B(4a, 4a). Also line 2bx + 3cy + 4d = 0 passes through A and B.

. ∴ d = 0 ............ (i)
or 2b . 4a + 3c . 4a + 4d = 0 2ab + 3ac + d = 0
a(2b + 3c) = 0 (Qd = 0)
⇒ 2b + 3c = 0 ............ (ii) On squaring equation (i) and (ii) and then adding, we get
d2 + (2b + 3c)2 = 0

868 Views

2.

If a circle passes through the point (a, b) and cuts the circle x2 +y2= 4 orthogonally, then the locus of its centre is

• 2ax +2by + (a2 +b2+4)=0

• 2ax +2by - (a2 +b2+4)=0

• 2ax -2by - (a2 +b2+4)=0

• 2ax -2by + (a2 +b2+4)=0

B.

2ax +2by - (a2 +b2+4)=0

Let the equation of circle is
x2 + y2 + 2gx + 2fy + c = 0
It cut the circle x2 + y2 = 4 orthogonally

if 2g.0 + 2f.0 = c - 4
⇒ c = 4
∴ Equation of circle is
x2 + y2 + 2gx + 2fy + 4 = 0
Q It passes through the point (a, b)
∴ a2 + b2 + 2ag + 2f + 4 = 0
Locus of centre (-g, -f) will be a2+ b2- 2xa - 2yb + 4 = 0
2ax + 2by - (a2 + b2 + 4) = 0

450 Views

3.

If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is

• x2 + y2- 2x +2y -23 = 0

• x2 - y2- 2x -2y -23 = 0

• x2 - y2- 2x -2y +23 = 0

• x2 + y2+ 2x +2y -23 = 0

A.

x2 + y2- 2x +2y -23 = 0

The lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 are diameters of circle.
On solving these equations we get x = 1, y = -1
Therefore the centre of circle = (1, -1) and circumference = 10 π
2πr = 10π
⇒ r = 5

∴ Equation of circle (x - x1 )2 + (y - y1 )2 = r2
(x - 1)2 + (y + 1)2 = 52
x2 + 1 - 2x + y2 + 2y + 1 = 25
x2 + y2 - 2x + 2y - 23 = 0

163 Views

4.

Area of the greatest rectangle that can be inscribed in the ellipse

• 2ab

• ab

• a/b

A.

2ab

Area of rectangle ABCD = (2acosθ)
(2bsinθ) = 2absin2θ
⇒ Area of greatest rectangle is equal to 2ab when sin2θ = 1.
178 Views

5.

The eccentricity of an ellipse, with its centre at the origin, is 1 /2 . If one of the directrices is x = 4, then the equation of the ellipse is

• 3x2 +4y2 = 1

• 3x2+ 4y2 = 12

• 4x2 +3y2 = 12

• 4x2+ 3y2 = 1

B.

3x2+ 4y2 = 12

Since the directrix is x = 4 then ellipse is parallel to X-axis.
⇒ a/e = 4
⇒ a = 4e = 4 x (1/2)
⇒ a = 2
Also we know that
b2 =a2 (1-e2)
b2 = 4(1-1/4) = 4 x 3/4
b2 = 3
therefore equation of ellipse is

311 Views

6.
• an ellipse

• a circle

• a straight line

• a parabola

C.

a straight line

As given  ⇒ distance of z from origin and point (0,1/3) is same hence z lies on the bisector of the line joining points (0, 0) and (0, 1/3). Hence z lies on a straight line.

208 Views

7.

Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is

• y2 – 4x + 2 = 0

• y2 + 4x + 2 = 0

• x2 + 4y + 2 =

• x2 – 4y + 2 = 0

A.

y2 – 4x + 2 = 0

P = (1, 0) Q = (h, k) such that
k2 = 8h
Let (α, β) be the midpoint of PQ

442 Views

8.

If |z2-1|=|z|2+1, then z lies on

• the real axis

• an ellipse

• a circle

• the imaginary axis

B.

an ellipse

Given that
|z2- 1| = |z|2+ 2
|z2 + (-1)| = |z2| + |-1|
It shows that the origin, -1 and z2 lies on a line and z2 and -1 lies on one side of the origin, therefore
z2 is a negative number. Hence z will be purely imaginary. So we can say that z lies on y-axis.

253 Views

9.

The locus of a point P (α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola

• an ellipse

• a circle

• a parabola

• a hyperbola

D.

a hyperbola

Tangent to the hyperbola

Given that y = αx + β is the tangent of hyperbola
⇒ m = α and a2 m2 – b2 = β2
∴ a2 α2 – b2 = β2
Locus is a2 x2 – y2 = b2 which is hyperbola.

219 Views

10.

A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is

• (x-p)2 = 4qy

• (x-q)2 = 4py

• (y-p)2 = 4qx

• (y-p)2 = 4px

A.

(x-p)2 = 4qy

In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B  is (x1 , y1 ).
Equation of circle in diameter form is (x - p)(x - x1 ) + (y - q)(y - y1 ) = 0
x2 - (p + x1 )x + px1 + y2 - (y1 + q)y + qy1 = 0
x2 - (p + x1 )x + y2 - (y1 + q)y + px1 + qy1 = 0
Since this circle touches X-axis

∴ y = 0
⇒ x2 - (p + x1 )x + px1 + qy1 = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x1 )2 = 4(px1 + qy1) p2 + x21 + 2px1
= 4px1 + 4qy1 x21 - 2px1 + p2 = 4qy1
Therefore the locus of point B is, (x - p)2 = 4qy

480 Views