Hi I'm attempting to change $9x^2-18x-y^2-8y-88=0$ to standard form. Here is what I've done: $$(9x^2 - 18x -1) - (y^2 + 8y +16) = 88+9+16$$ $$(3x-3)^2 - (y+4)^2 = 113$$ $$(3x-3)^2/113 - (y+4)^2/113 = 1$$ which would be for a horizontal hyperbola.
If I did the above correctly, I'm not sure how I would graph it with the $3x-3$. I also tried factoring the $9$ out first but I end up with another odd problem. Any advice would be greatly appreciated.
You can definitely divide out the nine as follows. $$\frac{(3x-3)^2}{113} - \frac{(y+4)^2}{113} = \frac{(x-1)^2}{113/9} - \frac{(y+4)^2}{113} = 1$$