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 ...
$\displaystyle x^2 + 8x + 16 - 3(y^2 + 2y + 1) = -4 + 16 - 3$
$\displaystyle (x + 4)^2 - 3(y + 1)^2 = 9$
$\displaystyle \frac{(x+4)^2}{9} - \frac{(y+1)^2}{3} = 1$
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