Question

The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is

• A. $\frac{x^2}{16}-\frac{y^2}{18}=1$
• B. $\frac{x^2}{36}-\frac{y^2}{27}=1$
• C. $\frac{x^2}{64}-\frac{y^2}{36}=1$
• D. $\frac{x^2}{36}-\frac{y^2}{64}=1$

Answer 1

Answer: (C)

Let equation of hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore e=\sqrt{1+\frac{b^2}{a^2}}\Rightarrow {\left(\frac{5}{4}\right)}^2$
$=\left(1+\frac{b^2}{a^2}\right)\left[\because e=\frac{5}{4}\right]$
$\Rightarrow \frac{b^2}{a^2}=\frac{9}{16}$
$\Rightarrow \frac{b}{a}=\frac{3}{4}$
We have, $\frac{2b^2}{a}=9$
$\Rightarrow 2\times \frac{3}{4}b=9$
$\Rightarrow b=6$ and $a=8$
$\therefore$ Equation of hyperbola is $\frac{x^2}{64}-\frac{y^2}{36}=1$