Question

Let $e$ be the eccentricity of hyperbola and $f(e)$ be the eccentricity of its conjugate hyperbola then $\underset{1}{\overset{\sqrt{2}}{\int }}(f(e)+f(f(e)))de$ is equal to

• A. 3
• B. 1
• C. $\frac{3}{2}$
• D. 2

Answer 1

Answer: (C)

Let eccentricity of conjugate hyperbola be e'
$\therefore \frac{1}{e^2}+\frac{1}{e^{\prime 2}}=1\Rightarrow \frac{1}{e^{\prime 2}}=1-\frac{1}{e^2}\Rightarrow e^{\prime }=\frac{e}{\sqrt{e^2-1}}$
$\therefore f(e)=\frac{e}{\sqrt{e^2-1}}$
and $f(f(e))=\frac{f(e)}{\sqrt{f^2(e)-1}}=\frac{\frac{e}{\sqrt{e^2-1}}}{\sqrt{\frac{e^2}{e^2-1}-1}}=e$
$\therefore$ Given integral $=\underset{1}{\overset{\sqrt{2}}{\int }}\left(\frac{e}{\sqrt{e^2-1}}+e\right)de={\left(\sqrt{e^2-1}+\frac{e^2}{2}\right)}_1^{\sqrt{2}}=(1+1)-\left(0+\frac{1}{2}\right)=\frac{3}{2}$