Question

What is the standard form of the equation of the parabola with a directrix at x=-16 and a focus at (12,-15)?

Answer 1

`x=1/56(y^2+30y+113)`

Explanation:

Given -
Directrix `x=-16)`
Focus `(12, -15)`

Its directrix is parallel to y-axis. So, this parabola opens to right.

The general form of the equation is

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

Where-

`h` x- coordinate of the vertex
`k` y-coordinate of the vertex
`a` is the distance between focus and vertex

Find the coordinates of the vertex.

Its y- coordinate is -15

Its x-coordinate is `(x_1+x_2)/2=(-16 + 12)/2=(-4)/2=-2`

Vertex is `(-2, -15)`

`a=14` distance between focus and vertex

Then -

`(y-(-15))^2=4xx14xx(x-(-2))`

`(y+15)^2=56(x+2)`
`y^2+30y+225=56x+112`

`56x+112=y^2+30y+225`
`56x=y^2+30y+225-112`
`56x=y^2+30y+113`
`x=1/56(y^2+30y+113)`