Question

What is the equation of a parabola with a focus at `(3, 2)` and a directrix at `y=-4`?

Answer 1

The equation of the parabola is `y=1/12(x-3)^2-1`

Explanation:

The focus is `F=(3,2)` and the directrix is `y=-4`

Any point `(x,y)` on the parabola is equidistant from the focus and the directrix.

`y+4=sqrt((x-3)^2+(y-2)^2)`

Squaring both sides

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

`y^2+8y+16=(x-3)^2+y^2-4y+4`

`12(y+1)=(x-3)^2`

`y+1=1/12(x-3)^2`

The equation is `y=1/12(x-3)^2-1`

graph{(y+1-1/12(x-3)^2)(y+4)=0 [-10, 10, -5, 5]}