Question

The directrix of the parabola `12(y+3)=(x-4)^2` has the equation `y=-6` What are the coordinates of the focus of the parabola?

Answer 1

Focus is `(4,0)`

Explanation:

`12(y+3)=(x-4)^2` or `y=1/12(x-4)^2-3`

In the equation `y=a(x-h)^2+k`, focus is `(h,k+1/(4a))` and directrix is `y=k-1/(4a)`. In case equation is `x=(y-k)^2+h`, focus is `(h+1/(4a),k)` and directrix is `x=h-1/(4a)`.

Here we have `a=1/12`, `h=4` and `k=-3` and `1/(4a)=3`. There directrix should be `y-3-3=-6` (which what is given),

vertex is `(4,-3)` and focus is `(4,-3+3)` or `(4,0)`.

The graph appears as shown below and vertex is equidistant from focus and directrix.

graph{(12y+36-(x-4)^2)((x-4)^2+y^2-0.05)(y+6)=0 [-6.83, 13.17, -6.68, 3.32]}