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AmberWscheid · Feb 6, 2012, 11:03 PM

In the problem is gives you the vertices of (4,1) and (4,9) and Foci of (4,0) and (4,10).
They want the standard form of the equation of the hyperbola. Thanks for your help!

corrigan · Feb 7, 2012, 09:50 AM

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.

AmberWscheid · Feb 8, 2012, 11:30 PM

Thank you so much! This explains a lot.