If the foci of a hyperbola are same as that of the ellipse and the eccentricity of the hyperbola is times the eccentricity of the ellipse, then the smaller focal distance of the point on the hyperbola, is equal to [2024]

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Solution

Q.
If the foci of a hyperbola are same as that of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}})$ on the hyperbola, is equal to [2024]

1 $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$

2 $14 \sqrt{\frac{2}{5}} - \frac{16}{5}$

3 $14 \sqrt{\frac{2}{5}} - \frac{4}{3}$

4 $7 \sqrt{\frac{2}{5}} + \frac{8}{3}$

Ans.
(1)

We have, $E : \frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$a = 5, b = 3$

foci : (0, $\pm$ 5e)

Since, for an ellipes, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow 9 = 25 (1 - e^{2})$

$e^{2} = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow e = \frac{4}{5}$

Eccentricity of hyperbola = $e' = \frac{15}{8} \times \frac{4}{5} = \frac{3}{2}$

Focus of hyperbola = (0, $\pm$ 5e) = (0, $\pm$ 4)

$\Rightarrow a \cdot e' = 4 \Rightarrow a = 4 \times \frac{2}{3} = \frac{8}{3}$

Since, $e'^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow \frac{9}{4} = 1 + \frac{b^{2}}{1} \times \frac{9}{64} \Rightarrow b^{2} = \frac{80}{9}$

$∴$ The equation of hyperbola $\frac{9 y^{2}}{64} - \frac{9 x^{2}}{80} = 1$

PS = $e' p M$

$= \frac{3}{2} \times (\frac{14}{3} \sqrt{\frac{2}{5}} - \frac{16}{9})$

$= 7 \sqrt{\frac{2}{5}} - \frac{8}{3}$ .

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