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  • If  and <img alt="\mathrm{e^{\prime}}" src="https://entrancecorner.oncodecogs.com/gif.l

If \mathrm{e} and \mathrm{e^{\prime}} be the eccentricities of a hyperbola and its conjugate, then \mathrm{\frac{1}{\mathrm{e}^2}+\frac{1}{\left(\mathrm{e}^{\prime}\right)^2}} is equal to

Option: 1

\mathrm{0}


Option: 2

\mathrm{1}


Option: 3

\mathrm{2}


Option: 4

\mathrm{\sqrt{2}}


Answers (1)

best_answer

Hyperbolas    \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}= \pm 1} are conjugate to each other.

Let  \mathrm{e}  be the eccentricity of   \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} and  \mathrm{e^{\prime}} be the eccentricity of

\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1} , then

     \mathrm{e^2=1+\frac{a^2}{b^2}}

     and  \mathrm{\left(e^{\prime}\right)^2=1+\frac{b^2}{a^2}}

     Now  \mathrm{\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1}

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