# 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

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

• Q2

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

Marks:1

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

• Q3

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

Marks:1

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

• Q4

Find the focus of parabola y2 = 8x.

Marks:1

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