Vertices, Directrices, and Foci. Oh my!

[delovely]

I really only need two problems checked & I need help solving another one...so here we go!

1. Vertex for a parabola with the equation (y-1)[sup:h928bvlv]2[/sup:h928bvlv]=4x

Is the vertex (0,1)? When you take the square root of 4x & y-1[sup:h928bvlv]2[/sup:h928bvlv], does the -1 stay negative so when you move it over it will become a positive one?

2. Directrix of the parabola with the equation y+3=1/10(x+2)[sup:h928bvlv]2[/sup:h928bvlv]

Is it -.5?

And I'm at a loss for this one:
3. The focus of the parabola with the equation (x-1)[sup:h928bvlv]2[/sup:h928bvlv]+32=8y? I think I need to get (x) by itself, right? Or is it y?should I divide the left side with 8? Help!! Ahhh!

 

[galactus]

And I'm at a loss for this one:
3. The focus of the parabola with the equation (x-1)[sup:btqys6tl]2[/sup:btqys6tl]+32=8y? I think I need to get (x) by itself, right? Or is it y?should I divide the left side with 8? Help!! Ahhh!

We have the form \(\displaystyle (x-h)^{2}=\pm 4p(y-k)\rightarrow (x-1)^{2}=8(y-4)\)

\(\displaystyle (x-1)^{2}=4p(y-4)\)

So, we can see p=2. So, the vertex and focus are 2 units apart.

The vertex is at (h,k), so the vertex is at V(1,4).

Since the equation is positive in 4p, it opens in the positive y direction.

Now, can you find the focus from here?.