If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that (1)/(e²

Exercise 7.3 · Q76

\dfrac{1}{e^2}+\dfrac{1}{e'^2}=1

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For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ : $e^2=1+\dfrac{b^2}{a^2}=\dfrac{a^2+b^2}{a^2} \Rightarrow \dfrac{1}{e^2}=\dfrac{a^2}{a^2+b^2}$ .

For its conjugate $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$ : $e'^2=1+\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2} \Rightarrow \dfrac{1}{e'^2}=\dfrac{b^2}{a^2+b^2}$ . …

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Q.If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that $\dfrac{1}{e^2}+\dfrac{1}{e'^2}=1$ .

Concept understanding — Focal Properties and Latus Rectum of a Hyperbola

Q.If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that 1e2+1e′2=1\dfrac{1}{e^2}+\dfrac{1}{e'^2}=1e21​+e′21​=1.

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Q.If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that $\dfrac{1}{e^2}+\dfrac{1}{e'^2}=1$ .

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