Ask Me Help Desk - Find the standard form of the hyperbola with the given characteristics?

Okay the first thing you need to notice is that this hyperbola opens up and down. The way I know that is because the vertices and the foci are all on the line $x=4$ . Now the equation for such a hyperbole looks like this:

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

And from that we can get the center, foci, and vertices. The center is at the point $(h,k)$ , the vertices are at $(h,k-a)$ and $(h,k+a)$ , and the foci are at $(h,k-\sqrt{a^2+b^2})$ and $(h,k+\sqrt{a^2+b^2})$ .

So what we need is to find $a$ , $b$ , $h$ , and $k$ . Well $h$ is easy, that's just $4$ . $k$ and $a$ aren't to bad either since $k-a = 1$ and $k+a=9$ . A little algebra, and we get $k=5$ and $a=4$ .

So all we need have left to find is $b$ . We now know what $k$ and $a$ are, and we know that $k-\sqrt{a^2+b^2}=0$ and $k+\sqrt{a^2+b^2}=10$ . By plugging our numbers in we get:
$5+\sqrt{4^2+b^2}=10$
$\Rightarrow \sqrt{16+b^2}=5$
$\Rightarrow 16+b^2=25$
$\Rightarrow b^2=9$ $\Rightarrow b=3$ . (b is always positive.)

So now we just plug everything in:

$\frac{(x-4)^2}{4^2} - \frac{(y-5)^2}{3^2} =1$

I'm sorry, I don't remember what "standard form" is, but from here you can manipulate it to anything you want. I hope this helps.