Ask Me Help Desk - What is the vertex 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 :)