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

Question

The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latusrectum $=9$ units and eccentricity $=\frac{5}{4}$, 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

Given, $\frac{2 b^2}{a}=9$ and $e=\frac{5}{4}$

$\Rightarrow \quad 1+\frac{b^2}{a^2}=\frac{25}{16} \Rightarrow \frac{b^2}{a^2}=\frac{9}{16}$

$\therefore \quad \frac{9}{2} \times \frac{1}{a}=\frac{9}{16} \Rightarrow a=8$

$\Rightarrow \quad b=6 \quad \therefore \frac{x^2}{64}-\frac{y^2}{36}=1$

So, The correct option of this question will be (C).