Question

What is the equation of the parabola with a focus at (3,6) and a directrix of y= 8?

Answer 1

`y=(-1/4)x^2+(6/4)x+(19/4)`

Explanation:

If the focus of a parabola is (3,6) and the directrix is y = 8, find the equation of the parabola.

Let ( x0 , y0 ) be any point on the parabola. First of all, finding the distance between (x0 , y0) and the focus. Then finding the distance between (x0 , y0) and directrix. Equating these two distance equations and the simplified equation in x0 and y0 is equation of the parabola.

The distance between (x0 , y0) and (3,6) is
`sqrt((x0-2)^2+(y0-5)^2`

The distance between (x0 , y0) and the directrix, y = 8 is | y0– 8|.

Equating the two distance expressions and square on both sides.
`sqrt((x0-3)^2+(y0-6)^2` = | y0– 8|.

`(x0-3)^2+(y0-6)^2` =`(y0-8)^2`

Simplifying and bringing all terms to one side:

`x0^2-6x0+4y0-19=0`

Write the equation with y0 on one side:
`y0=(-1/4)x0^2+(6/4)x0+(19/4)`

This equation in (x0 , y0) is true for all other values on the parabola and hence we can rewrite with (x , y).

So, the equation of the parabola with focus (3,6) and directrix is y = 8 is
`y=(-1/4)x^2+(6/4)x+(19/4)`