Question

The eccentricity of the conjugate hyperbola of $\frac{x^2}{16}-\frac{y^2}{9}=1$ is .

• A. $\frac{5}{4}$
• B. $\frac{25}{16}$
• C. $\frac{5}{3}$
• D. $\frac{25}{9}$

Answer 1

The correct option is C $\frac{5}{3}$

Let the eccentricity of the given hyperbola be e and that of its conjugate hyperbola be e'.
$\mathrm{Hence},e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}\mathrm{Also},\frac{1}{e^2}+\frac{1}{e{\text{'}}^2}=1\Rightarrow \frac{1}{e{\text{'}}^2}=1-\frac{16}{25}=\frac{9}{25}\Rightarrow e\text{'}=\frac{5}{3}$