# Hyperbola general form

• May 7th 2010, 08:19 PM
Anemori
Hyperbola general form
i have given equations:

$x^2-y=3$ <== parabola I already got this one.
$4x^2-y^2+2y=5$ ==> how to form into general form $\frac {(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$?

This i have this far:

$4x^2-(y^2+2y+1)=5+4-1$

$4(x-0)^2 - (y+2)^2 = 8$

$\frac {(x-0)^2}{2}-\frac{(y+2)^2}{8}=1$

$a=\sqrt(2), b=\sqrt(8), and (h,k) = (0,-2)$

Am I wrong or right? thanks in advance////
• May 7th 2010, 08:27 PM
sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...3baae6a6-1.gif
$4x^2 - y^2 + 2y - 1 = 5 - 1$
$4x^2 - ( y - 1)^2 = 4$
${x^2 - \frac{(y-1)^2}{2^2}}= 1$
• May 7th 2010, 09:32 PM
Anemori
Quote:

Originally Posted by sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...3baae6a6-1.gif
$4x^2 - y^2 + 2y - 1 = 5 - 1$
$4x^2 - ( y - 1)^2 = 4$
${x^2 - \frac{(y-1)^2}{2^2}}= 1$

Ah i see...

so a=1, b=2 and (h,k) = (0,1) right?
• May 7th 2010, 09:52 PM
sa-ri-ga-ma
Quote:

Originally Posted by Anemori
Ah i see...

so a=1, b=2 and (h,k) = (0,1) right?

Yes. You are right.