Conic Section

Sponsor Area

Question
CBSEENMA11015559

The ellipse x2+ 4y2= 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is

  • x2+ 16y2= 16 

  • x2+ 12y2= 16 

  • 4x2+ 48y2= 48 

  • 4x2+ 64y2= 48

Solution

B.

x2+ 12y2= 16 


straight x squared over 16 space plus straight y squared over straight b squared space equals space 1
It space passes space through space left parenthesis 2 comma 1 right parenthesis
So space 4 over 16 space plus 1 over straight b squared space equals space 1
1 over straight b squared space equals space 3 over 4
straight b squared space equals space 4 over 3
rightwards double arrow space straight x squared space plus space 12 straight y squared space equals space 16

Sponsor Area

Question
CBSEENMA11015567

A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi−major axis is 

  • 8/3

  • 2/3

  • 5/3

  • 4/3

Solution

A.

8/3

Major axis is along x-axis.
straight a over straight e minus space ae space equals space 4
straight a space open parentheses 2 minus 1 half close parentheses space equals 4
straight a space equals 8 over 3

Question
CBSEENMA11015568

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at 

  • (0, 2)

  • (1, 0)

  • (0,1)

  • (2,0)

Solution

B.

(1, 0)

vertex (0,1) 

Question
CBSEENMA11015586

Consider a family of circles which are passing through the point (-1, 1) and are tangent to x-axis. If (h, K) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interva

  • 0 < k < 1/2

  • k ≥ 1/2

  • – 1/2 ≤ k ≤ 1/2

  • k ≤ ½

Solution

B.

k ≥ 1/2

Equation of circle (x − h)2+ (y − k)2 = k2
It is passing through (− 1, 1) then
(− 1 − h)2+ (1 − k)2= k2
h2+ 2h − 2k + 2 = 0
D ≥ 0
2k − 1 ≥ 0 ⇒ k ≥ 1/2.