Question

If the hyperbola $\frac{x^2}{81}-\frac{y^2}{25}=1$, then find the value of eccentricity

• A. $\frac{\sqrt{96}}{9}$
• B. $\frac{\sqrt{105}}{5}$
• C. $\frac{\sqrt{96}}{5}$
• D. $\frac{\sqrt{105}}{9}$

Answer 1

The correct option is D $\frac{\sqrt{105}}{9}$

We have,

$\frac{x^2}{81}-\frac{y^2}{25}=1$

$\frac{x^2}{9^2}-\frac{y^2}{5^2}=1$

Here, $a=9$ and $b=5$

We know that

$e=\sqrt{1+\frac{b^2}{a^2}}$

Therefore,

$e=\sqrt{1+\frac{5^2}{9^2}}$

$e=\sqrt{1+\frac{25}{81}}$

$e=\sqrt{\frac{81+25}{81}}$

$e=\sqrt{\frac{105}{81}}$

$e=\frac{\sqrt{105}}{9}$

Hence, this is the answer.