How do you identify the conic section represented by the equation 9y^2-16x^2-72y-64x=64?

Dec 3, 2016

This is the equation of a hyperbola ${\left(y - 4\right)}^{2} / 16 - {\left(x + 2\right)}^{2} / 9 = 1$

Explanation:

Let rearrange the equation and complete the squares

$9 {y}^{2} - 72 y - 16 {x}^{2} - 64 x = 64$

$9 \left({y}^{2} - 8 y\right) - 16 \left({x}^{2} + 4 x\right) = 64$

$9 \left({y}^{2} - 8 y + 16\right) - 16 \left({x}^{2} + 4 x + 4\right) = 64 + 144 - 64$

$9 {\left(y - 4\right)}^{2} - 16 {\left(x + 2\right)}^{2} = 144$

$\frac{9}{144} {\left(y - 4\right)}^{2} - \frac{16}{144} {\left(x + 2\right)}^{2} = 1$

${\left(y - 4\right)}^{2} / 16 - {\left(x + 2\right)}^{2} / 9 = 1$

This is a hyperbola,

${\left(y - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $= \left(- 2 , 4\right)$

The equations of the asymptotes are

$y - 4 = \frac{4}{3} \left(x + 2\right)$

and $y - 4 = - \frac{4}{3} \left(x + 2\right)$

graph{(9y^2-16x^2-72y-64x-64)(y-4-4/3(x+2))(y-4+4/3(x+2))=0 [-83.36, 83.36, -41.6, 41.7]}