SOLUTION: How do i identify the vertex, focus, and directrix of y=1/12x^2

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do i identify the vertex, focus, and directrix of y=1/12x^2      Log On


   

Question 316878: How do i identify the vertex, focus, and directrix of y=1/12x^2
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is already in vertex form,
y=%281%2F12%29%28x-0%29%5E2%2B0
.
.
.
The vertex is (0,0).
.
.
.
The general equation for a parabola is,
4p%28y-k%29=%28x-h%29%5E2 where p is the distance from the vertex to the focus.
In your case,
12%28y-0%29=%28x-0%29%5E2
4p=12
p=3
The focus is located at (0,0)+(0,3)=(0,3)
.
.
.
The directrix is the same distance from the vertex but in the opposite direction.
y=-3
.
.
.