Let the eccentricity of an ellipse is reciprocal to that of the hyperbola If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

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Solution

Q.
Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1 .$ If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

Ans.
(2)

$We have, equation of hyperbola as \frac{x^{2}}{1 / 2} - \frac{y^{2}}{1 / 2} = 1$

$∴$ $e_{H} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{2}$

$So, e_{E} = \frac{1}{\sqrt{2}} (Given that e_{E} = \frac{1}{e_{H}})$

$∵$ $Ellipse and hyperbola intersect at 90 ° .$

$∴$ $They are confocal.$

$Now, foci of hyperbola = (1, 0)$

$∴$ $a e_{E} = 1 \Rightarrow a = \sqrt{2}, b = 1$

So, $Latus rectum of ellipse = \frac{2 b^{2}}{a} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$∴$ $Square of latus rectum is 2 .$

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