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What is the focus of 64y-4x^2?
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There is no focus. You have presented an algebraic expression. You will need an actual equation in order to identify a conic section and subsequently identify it's focus.
Oops my mistake *** The actual equation is 64y = 4x^2
That's better. You should have a "Standard Form" for a parabola. What is it? Usually, (y-k) = 4p(x-k)^2 -- or something like that.
(y-k)2=4p(x-h)
Whoops! Of course, I meant the Vertex Form. Do you have that?
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y = a(x – h) 2 + k ?
That's it. What does "a" mean in this form?
Just solved it, but thanks!
You're not going to share?
The answer was actually (0,4) :)
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