Question

How do you find the equation in standard form of the parabola whose focus is at (-2, 4) and whose directrix is at x = 4?

Answer 1

`x=-1/4y^2+2y-1`

Explanation:

Parabola is the locus of a point which moves whose distance from a given line called directrix and a point called focus is always constant.

Here we have directrix as `x=4` and focus is `(-2,4)`

Distance of `(x,y)` from `x=4` is `|x-4|`

and distance from focus `(-2,4)` is `sqrt((x+2)^2+(y-4)^2)`. Hence

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

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

or `x^2+4x+4+y^2-8y+16=x^2-8x+16`

or `y^2-8y+4x+4=0`

and for standard form it is `4x=-y^2+8y-4`

or `x=-1/4y^2+2y-1`
graph{x=-1/4y^2+2y-1 [-12.965, 7.035, -1, 9]}