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Conic Section

    • Question 1
    CBSEENMA11015559
    Wired Faculty App

    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
    Question 2
    CBSEENMA11015567
    Wired Faculty App

    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 3
    CBSEENMA11015568
    Wired Faculty App

    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 4
    CBSEENMA11015586
    Wired Faculty App

    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.

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