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Let e_{1} and e_{2} be the eccentricities of the ellipse, \frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b< 5) and the hyperbola, \frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1 respectively satisfying e_{1}e_{2}=1. If \alpha and \beta are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (\alpha ,\beta ) is equal to :
Option: 1 (8,12)
Option: 2 \left ( \frac{20}{3},12 \right )
Option: 3 \left ( \frac{24}{5},10 \right )
Option: 4 \left ( 8,10 \right )

Answers (1)

best_answer

\begin{aligned} &\text { For ellipse } \frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)\\ &\text { Let } e_{1} \text { is eccentricity of ellipse }\\ &\therefore \quad b^{2}=25\left(1-e_{1}^{2}\right) \ldots \ldots(1) \end{aligned}

For Hyperbola

\begin{aligned} &\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1\\ &\text { Let } \mathrm{e}_{2} \text { is eccentricity of hyperbola. }\\ &\therefore \quad \mathrm{b}^{2}=16\left(\mathrm{e}_{2}^{2}-1\right) \quad \ldots \ldots(2) \end{aligned}

From (1) and (2)

\begin{aligned} &25\left(1-\mathrm{e}_{1}^{2}\right)=16\left(\mathrm{e}_{2}^{2}-1\right)\\ &\text { Now } \mathrm{e}_{1} \cdot \mathrm{e}_{2}=1 \quad \text { (given) }\\ &\therefore \quad 25\left(1-\mathrm{e}_{1}^{2}\right)=16\left(\frac{1-\mathrm{e}_{1}^{2}}{\mathrm{e}_{1}^{2}}\right)\\ &\text { or }\\ &e_{1}=\frac{4}{5} \quad \therefore e_{2}=\frac{5}{4} \end{aligned}

distance beween foci = 2ae

\begin{aligned} &\therefore \text { distance for ellipse }=2 \times 5 \times \frac{4}{5}=8=\alpha\\ &\text { distance for hyperbola }=2 \times 4 \times \frac{5}{4}=10=\beta\\ &\therefore(\alpha, \beta) \equiv(8,10) \end{aligned}

 

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himanshu.meshram

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