Question

The equation of the hyperbola in the standard form (with transverse axis along the x-axis having the length of the latusrectum $=9$ unit 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

The correct option is C $\frac{x^2}{64}-\frac{y^2}{36}=1$

Length of latusrectum $=9=\frac{2b^2}{a}$

$\Rightarrow b^2=\frac{9a}{2}$ ..... (i)

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

$\because 1+\frac{b^2}{a^2}=\frac{25}{16}$

$\therefore 1+\frac{9a}{2a^2}=\frac{25}{16}$ [from Eq. (i)]

$\Rightarrow \frac{9}{2a}=\frac{9}{16}\Rightarrow a=8$

On putting the value of a in Eq. (i), we get

$b^2=\frac{9\times 8}{2}=36\Rightarrow b=6$

$\therefore$ Equation of hyperbola is

$\frac{x^2}{8^2}-\frac{y^2}{6^2}=1\Rightarrow \frac{x^2}{64}-\frac{y^2}{36}=1$