An ellipse $E : \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and hyperbola $H$ whose length of tansverse axis is equal to semi minor axis of ellipse $E$ , are confocal. Point $P$ is the common point for $E$ and $H$ and $M_{1}$ and $M_{2}$ are the foot of perpendiculars from $P$ to corresponding directrix of $E$ and $H$ respectively then distance between $M_{1}$ and $M_{2}$ is :

7

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\frac{11}{5}

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5

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14

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Solution

The correct option is C $7$
For confocal ellipse and hyperbola,
$a e_{1} = A e_{2}$
Here $a$ and $e_{1}$ are semimajor axis and eccenticity of ellipse and A and $e_{2}$ are for hyperbola.
$\Rightarrow 5 \times \frac{3}{5} = 2 \times e_{2} (∴ 2 A = 4)$
$\Rightarrow e_{2} = \frac{3}{2}$
$∵ M_{1} M_{2} = (P M_{1} + P M_{2})$
$=$ Distance between their directrices.
$= \frac{a}{e_{1}} - \frac{A}{e_{2}} = \frac{5}{(3 / 5)} - \frac{2}{(3 / 2)} = \frac{25}{3} - \frac{4}{3} = 7$

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The correct option is C 7For confocal ellipse and hyperbola,ae1=Ae2Here a and e1 are semimajor axis and eccenticity of ellipse and A and e2 are for hyperbola.⇒5×35=2×e2(∴2A=4)⇒e2=32∵M1M2=(PM1+PM2) = Distance between their directrices. =ae1−Ae2=5(3/5)−2(3/2)=253−43=7