Question

How do you find the standard form of the equation of the parabola with a focus at `(0,-8)` and a directrix at `y=8`?

Answer 1

The equation of parabola is `y= -1/32 x^2`

Explanation:

The equation of parabola in standard form is

`y=a(x-h)^2+k ; (h,k)` being vertex.

Focus is at `0, -8` and directrix is `y=8`

Vertex is at midway between focus and directrix. Therefore vertex

is at `(0,(8-8)/2) or (0,0)`

The distance of focus from vertex is `d=0-(-8)=8`.

The focus is below the vertex , so the parabola opens downwards

and `a` is negative. `|a|= 1/(4d) =1/(4*8)=1/32 :. a = -1/32`

The equation of parabola is `y=-1/32(x-0)^2+0` or

`y= -1/32 x^2`

graph{-1/32x^2 [-160, 160, -80, 80]}