# How do you write the standard form of the hyperbola -16x^2+9y^2+32x+144y-16=0?

We can rewrite this as follows

$16 {x}^{2} - 9 {y}^{2} - 32 x - 144 y + 16 = 0$ (Multiply with $- 1$ both sides)

$16 \cdot \left({x}^{2} - 2 x\right) - 9 \left({y}^{2} + 16 y\right) + 16 = 0$

$16 \cdot \left({x}^{2} - 2 x + 1\right) - 9 \left({y}^{2} + 16 + 64\right) + \left(16 - 16 + 576\right) = 0$

$16 \cdot {\left(x - 1\right)}^{2} - 9 \cdot {\left(y + 8\right)}^{2} = - 576$

${\left(y + 8\right)}^{2} / 64 - {\left(x - 1\right)}^{2} / 36 = 1$

${\left(\frac{y + 8}{8}\right)}^{2} - {\left(\frac{x - 1}{6}\right)}^{2} = 1$

Finally the standard form for the hyperbola is

${\left(y - \left(- 8\right)\right)}^{2} / {8}^{2} - {\left(x - 1\right)}^{2} / {6}^{2} = 1$

If we plot it on the cartesian plane it is