Conic Sections - Online Test

Q1. The equation  represents a
Answer : Option C
Explaination / Solution:

On factorizing the given equation we get x-4y=0 and x-3y=0, which represents a pair of intersecting straight lines. Since the product of their slopes is not equal to -1, they are non perpendicular.

Q2. The equation  represents
Answer : Option B
Explaination / Solution:

On factorizing the given equation we get y+x-1=0 and y-x+1=0, which clearly represent a pair of straight lines which are intersecting.

Q3. The length of latus rectum of an ellipse is one-third of its major axis. Its eccentricity would be
Answer : Option D
Explaination / Solution:

= 2a/3

Hence  = 1/3. Hence e2 = 1- = 1 - (1/3) = 2/3

Therefore e = 


Q4. Three normals to the parabola  are drawn through a point (c, 0) then
Answer : Option D
Explaination / Solution:

The equation of the normal to a parabola y2 = 4ax is

y = mx -2am - am3

Hence the equation of the normal to the given parabola y2 = x is 

mx - 

Since it passes throught (c,0)

mc - = 0

on solving we  get 

m = 0 or m2 = 4(c-1/2)

If m = 0 then the equation of the normal is y = 0

If m2 0, then  4(c-1/2) 0

Hence c-1/2 0 or c > 1/2


Q5. S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
Answer : Option B
Explaination / Solution:

s = (ae,0) and T = (-ae,0) and B = (0,b)

Since it is an equilateral triangle, ST2 = TB2

This implies 4a2e2 = a2e2 + b2

3a2e2 = b

3a2e2 = a2(1 - e2)

3e2  = 1 - e2

Therefore e = 1/2


Q6.

The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle  is


Answer : Option A
Explaination / Solution:

Since the circle passes through (0,0) the equation reduces to 

c= 0 -----(1)

Since it passes through (1,0),

1 + 2g + c = 0

This implies g = -1/2

Since the circle touches the circle x2 + y2 = 9, their radii should be equal

2 = 3

Substituting the values and simplifying we get f = 

Hence the centre is (1/2, )


Q7. The number of tangents to the circle through the point ( - 1, 2) is
Answer : Option A
Explaination / Solution:

The given equation of the circle can be written as 

(x-1)2 - 1 + (y- 2)2 - 4 = 0

(x - 1)2 + (y + 2)2 = 5

This implies the radius is  and the centre is (1,-2)

The given point is (-1,2)

The distance between the centre of the  circle and the given point is

 = 

Sice this is greater than the radius, the point lies outside the circle. Hence two tangents can be drawn.


Q8. The slope of the tangent at the point (a, a) of the circle  is
Answer : Option C
Explaination / Solution:

the equation of the tangent is x+y=a. Hence its slope is -1

Q9. The equation of the normal to the parabola  having slope 1 is
Answer : Option C
Explaination / Solution:

slope form of normal is y=mx-2am-am3

for the given parabola a = 2 and m = 1

therefore y = x -4 -2 

i.e; x - y - 6 = 0


Q10. Which of the following lines is a normal to the circle 
Answer : Option A
Explaination / Solution:


comparing the above equation with we get center as (1,2)

putting (1,2) in L.H.S of above line ,we get 1+2=3=R.H.S

=> it satisfies center satisfies the equation of line or the line passes through center

and the line passing through the centre is normal.