# finding hyperbola vertices, ect.

• December 6th 2009, 07:03 PM
RenSully
finding hyperbola vertices, ect.
Find the vertices, foci, and the equations of the asymptotes of the hyperbola.
http://college.hmco.com/blackboard/t...s/hl2202_1.gif http://cengage.blackboard.com/images/spacer.gif

I am having trouble with this because I can't seem to find the right "a"

When I put the equation in standard form would "a" be 1 or 9/1? and then when I get the vertices I use the equation: (h+a,k) and (h-a,k). So it would be (0+1,0) or (0+9,0) and neither of these are the correct answer.
• December 6th 2009, 07:10 PM
VonNemo19
Quote:

Originally Posted by RenSully
Find the vertices, foci, and the equations of the asymptotes of the hyperbola.
http://college.hmco.com/blackboard/t...s/hl2202_1.gif http://cengage.blackboard.com/images/spacer.gif

I am having trouble with this because I can't seem to find the right "a"

When I put the equation in standard form would "a" be 1 or 9/1? and then when I get the vertices I use the equation: (h+a,k) and (h-a,k). So it would be (0+1,0) or (0+9,0) and neither of these are the correct answer.

So you have
$9x^2-\frac{16}{9}y^2=1$

Then

$\frac{x^2}{\frac{1}{9}}-\frac{y^2}{\frac{9}{16}}=1$
and

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

Does this make any sense to you?
• December 6th 2009, 07:22 PM
RenSully
yes, absolutely! Thanks for reminding me. I didn't even think of it that way. I took college algebra 5 years ago so I am very rusty.