# How do you find the standard form of 9x^2-16y^2-90x+32y+65=0?

Apr 15, 2016

#### Answer:

${\left(x - 5\right)}^{2} / {4}^{2} - {\left(y - 1\right)}^{2} / {3}^{2} = 1$

#### Explanation:

$9 {x}^{2} - 90 x = 9 {\left(x - 5\right)}^{2} - 225$.
$- 19 {y}^{2} + 32 y = - 16 {\left(y - 1\right)}^{2} + 16$.

So, the given equation becomes
$9 {\left(x - 5\right)}^{2} - 16 {\left(y - 1\right)}^{2} = 144$
Divide by 144.
${\left(x - 5\right)}^{2} / {4}^{2} - {\left(y - 1\right)}^{2} / {3}^{2} = 1$.

This equation represents a hyper bola with semi-major axis a = 4 and semi-transverse axis b = 3.

The eccentricity of the hyperbola is 1.25, using the relation ${b}^{2} = {a}^{2} \left({e}^{2} - 1\right)$

The asymptotes are $y = \pm \left(\frac{b}{a}\right) x = \pm \frac{3}{4} x$.