Question

Eccentricity of the hyperbola, conjugate to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, is

• A. $\frac{9}{5}$
• B. $\frac{5}{3}$
• C. $\frac{5}{4}$
• D. $\frac{9}{4}$

Answer 1

Answer: (B)

$\frac{5}{3}$

Eccentricity of given hyperbola $=\sqrt{1+\frac{9}{16}}=\frac{5}{4}=e_1$ (say)
If eccentricity of conjugate hyperbola is $e_2$
$\therefore \frac{1}{e_1^2}+\frac{1}{e_2^2}=1$
$\Rightarrow \frac{16}{25}+\frac{1}{e_2^2}=1\Rightarrow \frac{1}{e_2^2}=1-\frac{16}{25}=\frac{9}{25}$
$\therefore e_2=\frac{5}{3}$
Hence, option 'B' is correct