Question

What is the equation of a parabola with a vertex at (2,3) and a focus at (6,3)?

Answer 1

`(y-3)^2 = 16 (x-2)` is the equation of the parabola.

Explanation:

Whenever vertex(h,k) is known to us, we must preferably use the vertex form of the parabola:
(y−k)2=4a(x−h) for horizontal parabola
(x−h)2=4a(y−k) for veretical parabola

+ve when focus is above the vertex(vertical parabola) or when focus is to the right of vertex(horizontal parabola)

-ve when focus is below the vertex(vertical parabola) or when focus is to the left of vertex(horizontal parabola)

Given Vertex (2,3) and focus (6,3)

It can be easily noticed that focus and vertex lie on the same horizontal line y = 3

Obviously, the axis of symmetry is a horizontal line ( a line perpendicular to y-axis). Also, the focus lies to the right of the vertex so the parabola will open up rightward.

`(y-k)^2 = 4 a (x-h)`

`a = 6 - 2 = 4` as y coordinates are the same.

Since the focus lies to the left of vertex, a = 4

`(y-3)^2 = 4 * 4 * (x - 2)`

`(y-3)^2 = 16 (x-2)` is the equation of the parabola.

Answer 2

The equation of parabola is `(y-3)^2=16(x-2)`

Explanation:

Focus is at `(6,3)`and vertex is at `(2,3);h=2,k=3`.

Since focus is at right of vertex, parabola opens right ward

and `a` is positive. The equation of right opened parabola is

`(y-k)^2=4a(x-h) ; (h.k) ;` being vertex and focus is at

`(h+a,k):. 2+a=6 :. a= 6-2=4` . Hence equation of

parabola is `(y-3)^2=4*4(x-2) or (y-3)^2=16(x-2)`

graph{(y-3)^2=16(x-2) [-80, 80, -40, 40]} [Ans]