Question

Find the eccentricity of the hyperbola $9x^2-16y^2+72x-32y-16=0$.

Answer 1

Answer: 1.25

$9x^2-16y^2+72x-32y-16=0$
$9(x+4{)}^2-16(y+1{)}^2=144$
$\frac{(x+y{)}^2}{16}-\frac{(y+1{)}^2}{9}=1$
$e=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$