# Find vertex, focus, and directrix of parabola: y^2+12y+16x+68=0

#### stephensnoah1

##### New member
Find the vertex, focus, and directrix of the following parabola:

y^2+12y+16x+68=0

The form we have been using is (y-k)^2=4p(x-h)

Any explanation would help too, I'm really stuck on this one.

Thank you!

#### Olinguito

##### Well-known member
Hint: Rewrite the equation of the parabola as
$$16x\ =\ 68-12y-y^2\ =\ 104-(y+6)^2$$
whence
$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$
in the form you have been using.