Let e be the eccentricity of a hyperbola and f ( e ) be the eccentricity of its conjugate hyperbola, then ? ( f ( e ) + f ( f ( e ) ) ) d e = g ( e ) and g ( 2 ) = 2 , then g ( e ) =

Question

Let $e$ be the eccentricity of a hyperbola and $f (e)$ be the eccentricity of its conjugate hyperbola, then $\int (f (e) + f (f (e))) d e = g (e)$ and $g (\sqrt{2}) = 2$ , then $g (e) =$

Smart Achievers · Accepted Answer

(3)

$Let eccentricity of conjugate hyperbola be e'$

$∴$ $\frac{1}{e^{2}} + \frac{1}{e'^{2}} = 1 \Rightarrow \frac{1}{e'^{2}} = 1 - \frac{1}{e^{2}} \Rightarrow e' = \frac{e}{\sqrt{e^{2} - 1}}$

$∴$ $f (e) = \frac{e}{\sqrt{e^{2} - 1}}$

$and f (f (e)) = \frac{f (e)}{\sqrt{f^{2} (e) - 1}} = \frac{\frac{e}{\sqrt{e^{2} - 1}}}{\sqrt{\frac{e^{2}}{e^{2} - 1} - 1}} = e$

$∴$ $Given integral = \int (\frac{e}{\sqrt{e^{2} - 1}} + e) d e$ $= (\sqrt{e^{2} - 1} + \frac{e^{2}}{2})$

$g (e) = \sqrt{e^{2} - 1} + \frac{e^{2}}{2} + C$

$g (\sqrt{2}) = 1 + 1 + C$ $\Rightarrow C = 0$

$g (e) = \sqrt{e^{2} - 1} + \frac{e^{2}}{2}$

Solution

Q.
Let $e$ be the eccentricity of a hyperbola and $f (e)$ be the eccentricity of its conjugate hyperbola, then $\int (f (e) + f (f (e))) d e = g (e)$ and $g (\sqrt{2}) = 2$ , then $g (e) =$

1 $\frac{1}{2} \sqrt{e^{2} - 1} + \frac{e^{2}}{2}$

2 $\frac{1}{2} \sqrt{e^{2} + 1} + \frac{e^{2}}{2}$

3 $\sqrt{e^{2} - 1} + \frac{e^{2}}{2}$

4 $\sqrt{e^{2} + 1} + \frac{e^{2}}{2}$

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Solution

Q. Let e be the eccentricity of a hyperbola and f(e) be the eccentricity of its conjugate hyperbola, then ∫(f(e)+f(f(e)))de=g(e) and g(2)=2, then g(e)=

1 12e2-1+e22

2 12e2+1+e22

3 e2-1+e22

4 e2+1+e22