Question

If sec $\theta$ is the eccentricity of a hyperbola then the eccentricity of the conjugate hyberpola is

• A. tan θ
• B. cot θ
• C. cos θ
• D. cosec θ

Answer 1

The correct option is D

cosec θ

e = $\sqrt{1+\frac{b^2}{a^2}}=sec\theta$ $\Rightarrow$ $\frac{b^2}{a^2}$ = $se{c}^2\theta -1$ = $Ta{n}^2\theta$ $\Rightarrow \frac{a^2}{b^2}=co{t}^2\theta$

$\Rightarrow$ $e^1=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+co{t}^2\theta }=cosec\theta$