Question
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,prove that $\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1$.

Answer

Let e be the eccentricity of a hyperbola
$\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$
$e =\frac{\sqrt{ a ^2+ b ^2}}{ a }$
$e ^2=\frac{ a ^2+ b ^2}{ a ^20}$
$\frac{1}{ e ^2}=\frac{ a ^2}{ a ^2+ b ^2}$
$\ldots$..(i)
Also, $e ^{\prime}$ is the eccentricity of conjugate
hyperbola $\frac{y^2}{ b ^2}-\frac{x^2}{ a ^2}=1$
$e ^{\prime}=\frac{\sqrt{ b ^2+ a ^2}}{ b }$
$\left( e ^{\prime}\right)^2=\frac{ b ^2+ a ^2}{ b ^2}$
$\frac{1}{\left(e^{\prime}\right)^2}=\frac{b^2}{b^2+a^2}$
...(ii)
Adding (i) and (ii), we get
$\frac{1}{ e ^2}+\frac{1}{\left( e ^{\prime}\right)^2}=\frac{ a ^2}{ a ^2+ b ^2}+\frac{ b ^2}{ a ^2+ b ^2}=\frac{ a ^2+ b ^2}{ a ^2+ b ^2}$
$\frac{1}{ e ^2}+\frac{1}{\left( e ^{\prime}\right)^2}=1$

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