Parabola

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  • Q1

    Find the equation of the parabola with vertex at origin. The parabola is symmetric with respect to y-axis and passes through (3,–5).

    Marks:2
    Answer:

    The vertex of the parabola is at origin and it is
    symmetric with respect to y-axis, i.e., axis of the
    parabola lies along y-axis.
    The parabola passes through (3, –5), a point in the fourth quadrant. Therefore, the parabola lies in the fourth quadrant and its standard equation is
    x2 = – 4ay. Point (3, –5) lies on this curve then
    (3)2 = –4a (–5)

    9 = 20a
    a = (9/20)

    Thus, the equation of the parabola is

    x2 = – 4(9/20)y
    = – (9/5)y

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  • Q2

    Find the directrix for the conic section represented by the equation of 7x2 = 20y.

    Marks:1
    Answer:

    On comparing 7x2 = 20y with x2 = 4ay, we get
    4a = 20/7
    a = 5/7
    So, directrix of parabola is y + (5/7) = 0.

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  • Q3

    Find the length of latus rectum of parabola y2 = 20x.

    Marks:1
    Answer:

    On comparing y2 = 20x with y2 = 4ax, we get 4a = 20
    a = 5
    So, the length of latus rectum is 4(5) =20.

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  • Q4

    Find the focus of parabola y2 = 8x.

    Marks:1
    Answer:

    On comparing y2 = 8x with y2 = 4ax, we get
    4a = 8a = 2
    So, the coordinates of focus are (2, 0).

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  • Q5

    What is the equation of directrix of parabola y2= –4ax?

    Marks:1
    Answer:

    Equation of directrix of parabola y2 = –4ax is x – a =0.

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