Question

If the eccentricity of the hyperbola $9x^2-16y^2+72x-32y-16=0$ is $e$ then $4e=$

Answer 1

Answer: 5

$9x^2-16y^2+72x-32y-16=0$
$9{\left(x+4\right)}^2-16{\left(y+1\right)}^2=144$
$\frac{{\left(x+y\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1$
$e=\frac{\sqrt{a^2+b^2}}{a};a^2=16\&b^2=9.$
$e=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Hence, $4e=5$ .