Question

For the hyperbola $\frac{x^2}{100}-\frac{y^2}{25}=1$, prove that eccentricity $=\frac{\sqrt{5}}{2}$.

Answer 1

Q. For the Hyperbola $\frac{x^2}{100}-\frac{y^2}{25}=1$

Prove that eccentricity $=\frac{\sqrt{5}}{2}$

Sol${}^n\Rightarrow$ Let compare the equation with

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

we get, $a^2=100$ & $b^2=25$

Now, eccentricity (e) is given by,

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

$e=\sqrt{1+\frac{25}{100}}$

$e=\sqrt{\frac{125}{100}}$

$e=\sqrt{\frac{5}{4}}$

$e=\sqrt{\frac{5}{2}}$