If eccentricities of two hyperbolas x2-a2-y2-b2-1 and x2-a2-y2-b2-1 are e1 and e2 respectively then prove that 1-e21-1-e22-1-

The eccentricity e_1 of the given hyperbola is obtained from b^2 = a^2(e_1^2 1) … (1) The eccentricity e_2 of the conjugate hyperbola is given ...

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Standard XII

Mathematics

Conjugate Hyperbola

If eccentrici...

Question

If eccentricities of two hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ are $e_{1}$ and $e_{2}$ respectively then prove that $\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = 1$ .

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Solution

The eccentricity $e_{1}$ of the given hyperbola is obtained from
$b^{2} = a^{2} (e_{1}^{2} - 1) \dots (1)$
The eccentricity $e_{2}$ of the conjugate hyperbola is given by
$a^{2} = b^{2} (e_{2}^{2} - 1) \dots (2)$
Multiply (1) and (2), we get,
$1 = (e_{1}^{2} - 1) (e_{2}^{2} - 1)$
$⟹ 0 = e_{1}^{2} e_{2}^{2} - e_{1}^{2} - e_{2}^{2}$
$⟹ e_{1}^{- 2} + e_{2}^{- 2} = 1$
Hence proved.

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If eccentricities of two hyperbolas x2a2−y2b2=1 and x2a2−y2b2=−1 are e1 and e2 respectively then prove that 1e21+1e22=1.

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