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]

Question

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]

Smart Achievers · Accepted Answer

(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 = $e' p M$

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

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

Solution

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

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Solution

Q. If the foci of a hyperbola are same as that of the ellipse x29+y225=1 and the eccentricity of the hyperbola is 158 times the eccentricity of the ellipse, then the smaller focal distance of the point (2,14325) on the hyperbola, is equal to [2024]

1 725–83

2 1425–165

3 1425–43

4 725+83

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