What is the focus and vertices of 2x^2 4=4x y^2

I'm assuming that you mean 2x^2 + 4 = 4x +y^2 .
The first thing we need to do is put this in a form we can use. I can tell you from experience that this is an up-down hyperbola, because it is of the form ay^2 - bx^2 =1 . A left-right hyperbola would be of the form ax^2 - by^2 =1 and the way you would get the foci and vertices is a little different, you would be swapping some of the x and y values. But right now we are focusing on your problem.

So we need to make 2x^2 + 4 = 4x +y^2 into something that looks more like:

\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} =1

Then our center will be at the point (h,k) ,
vertices will be at the points (h,k+a) and (h,k-a) ,
and the foci will be at the points (h,k+\sqrt{a^2 +b^2}) and (h,k+\sqrt{a^2 +b^2}) .
You never said that you needed the asymptotes, but we'll do those too. They are just the lines with the equations

y-a=\frac{a}{b}(x-h) and y-a=-\frac{a}{b}(x-h) .

Okay, enough abstract letters. Let's get to the actual problem.

2x^2 + 4 = 4x +y^2

\Rightarrow x^2 + 2 = 2x +\frac{y^2}{2} (I just divided both sides by 2).

\Rightarrow 1 = -x^2 - 1+2x +\frac{y^2}{2}

\Rightarrow 1 =\frac{y^2}{2} -x^2+2x - 1

\Rightarrow 1 =\frac{y^2}{2} -(x^2-2x + 1)

\Rightarrow 1 =\frac{y^2}{2} -(x-1)^2

\Rightarrow 1 =\frac{(y-0)^2}{2} -\frac{(x-1)^2}{1} .

So we have a = sqrt{2} , b=1 , h=0 , and k =1. Now we just plug in the numbers and get:

center: (0,1)

vertices: (0,1+sqrt{2}) and (0,1-sqrt{2})

foci: (0,1+sqrt{5}) and (0,1-sqrt{5})

asymptotes: y=\sqrt{2}(x-1) and y=-\sqrt{2}(x-1)

I hope this helps :)