# how do i graph a hyperbola from the following equation?

• December 3rd 2012, 03:45 PM
alyse2526
how do i graph a hyperbola from the following equation?
x^2-3y^2+8x-6y+4=0

how do i find the center,transverse axis, vertices, foci and asymptotes from that equation??
• December 3rd 2012, 04:29 PM
Prove It
Re: how do i graph a hyperbola from the following equation?
Move all the x terms together, the y terms together, and put the equation into the standard form by completing the square on the x terms and the y terms.
• December 3rd 2012, 04:44 PM
skeeter
Re: how do i graph a hyperbola from the following equation?
Quote:

Originally Posted by alyse2526
x^2-3y^2+8x-6y+4=0

how do i find the center,transverse axis, vertices, foci and asymptotes from that equation??

start by completing the square in x and y ...

$x^2 + 8x + 16 - 3(y^2 + 2y + 1) = -4 + 16 - 3$

$(x + 4)^2 - 3(y + 1)^2 = 9$

$\frac{(x+4)^2}{9} - \frac{(y+1)^2}{3} = 1$

now you have the hyperbola in the form ...

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