Question

What is the equation of the hyperbola having latus rectum and eccentricity 8 and $\frac{3}{\sqrt{5}}$ respectively?

• A. $\frac{x^2}{25}-\frac{y^2}{20}=1$
• B. $\frac{x^2}{40}-\frac{y^2}{20}=1$
• C. $\frac{x^2}{40}-\frac{y^2}{30}=1$
• D. $\frac{x^2}{30}-\frac{y^2}{25}=1$

Answer 1

The correct option is A $\frac{x^2}{25}-\frac{y^2}{20}=1$

Length of latusrectum of the hyperbola $\frac{2b^2}{a}=8$ (Given) ....$\left(1\right)$

Or $b^2=4a$

Eccentricity of the hyperbola $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{3}{\sqrt{5}}$

$\Rightarrow e^2=1+\frac{b^2}{a^2}=\frac{9}{5}$

$\Rightarrow \frac{b^2}{a^2}=\frac{4}{5}$ ...$\left(2\right)$

Putting value of $b^2$ from equation $\left(1\right)$ into equation $\left(2\right)$ we get,

$\Rightarrow \frac{4a}{a^2}=\frac{4}{5}$

Or $a=5$ as $a\ne 0$

$\Rightarrow a^2=25$

$\Rightarrow b^2=4a=4\times 5=20$

Hence the equation of the given hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}=1$.

So correct option is $A$.