1. Tardigrade
  2. Question
  3. Mathematics
  4. If the eccentricity of the hyperbola 9x2-16y2+72x-32y-16=0 is e then 4e=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

NTA AbhyasNTA Abhyas 2022

Solution:

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