x^2-3y^2+8x-6y+4=0

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

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- Dec 3rd 2012, 03:45 PMalyse2526how 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?? - Dec 3rd 2012, 04:29 PMProve ItRe: 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.

- Dec 3rd 2012, 04:44 PMskeeterRe: how do i graph a hyperbola from the following equation?
start by completing the square in x and y ...

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

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

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

now you have the hyperbola in the form ...

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

go to this page and see how everything is related ...

Hyperbola Formulas