Hyperbola equation converted to standard form

[Louise Johnson]

Hello this is my poor attempt at converting to standard form. Any kind of help would be a major help.
Thank you
Louise

3x²-y²+6x+4y-10=0

(x+3) ² _ (x+2) ² =1
3/23 .. ..... 23


When graphed it would be a hyperbola with:
center -3,-2

Vertices
-2 sqrt 3/23 , -2
-3 sqrt 3/23 , -2

 

[skeeter]

\(\displaystyle \L 3x^2 - y^2 + 6x + 4y - 10 = 0\)

\(\displaystyle \L 3(x^2 + 2x) - (y^2 - 4y) = 10\)

\(\displaystyle \L 3(x^2 + 2x + 1) - (y^2 - 4y + 4) = 10 + 3 - 4\)

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

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

 

[Louise Johnson]

:wink: Thank you skeeter !!! I knew it couldn't be as complicated as I made it. I totally see where I screwed up. My basic factoring is crap and when I took out the initial 3 I left the 6 as it was instead of factoring it by 3 as well.

Still gonna work throught it some more and make sure I graph it properly.
THANKYOU!!!!!!!
Louise