Attempting to put the following conic into standard form: $y^2-2x^2+8y-8x-4=0$

Question

Put the following conic into standard form: $y^2-2x^2+8y-8x-4=0$

I ended up with $$ -\frac {(x-2)^2}{12} + \frac{(y+8)^2}{24} = 1, $$ but I'm not sure if this is right.

Answer 1

$$ y^2 -2x^2 + 8y - 8x - 4 = 0 \implies (y^2 + 8y + 16 - 16) - 2(x^2 + 4x + 4 - 4) - 4 = 0 \implies \\ (y + 4)^2 - 2(x+2)^2 - 16 + 8 - 4 = 0 \implies (y+4)^2 - 2(x+2)^2 = 12 \implies \\ -\frac {(x+2)^2}6 + \frac {(y+4)^2}{12} = 1 $$