write the equation of a parabola in vertex form given vertex (3,5), and a point on the parabola (1,-3).?

Question

Please show steps, thanks.
I got the first step:
y= a(x-3)^2+5
then i sub in the point:
-3=a(1-3)^2+5 ,after this im stumped of what to do

i never asked for focus and directrix -- this is toward the 3rd answer,

i need to find the "a" value

Answer 1

Solve for a,
a = -8/4 = -2
y = -2(x-3)^2 + 5

Answer 2

I like (x-3)^2=2p(y-5)
Then (1-3)^2 = 2p(-3-5)
4 = 2p(-8)
2p = -1/2
So equation is (x-3)^2 = -.5(y-5)
2p is the distance between the focus and the directrix. The vertex is midway between the focus and the directrix and vertex and focus lie on axis of symmetry. This form gives you a lot of data about the parabola.

Answer 3

You can do it, just figure it out...

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