SOLUTION: How do I find the vertex and focus and directrix of y=1/2x^2

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Question 1069405: How do I find the vertex and focus and directrix of y=1/2x^2
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
y=%281%2F2%29x%5E2

2y=x%5E2

highlight%282%28y-0%29=%28x-0%29%5E2%29
Vertex is (0,0).

The parabola has a focus above the vertex and directrix below the vertex.
Compare to 4py=x%5E2 for p the distance from vertex to focus.

4p=2
p=2%2F4
highlight_green%28p=1%2F2%29



The focus is 1%2F2 unit from the vertex (0,0).
FOCUS: (0, 1/2).
DIRECTRIX: y=-1%2F2.