QJEEM20Ma5b2c: JEE Main 2020 Mathematics - Class 11 JEEM20Ma5b2c Solution: For some , if the eccentricity of the hyperbola, x 2 –y 2 sec 2 …

Question

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For some

\theta \in \left( {0,{\pi \over 2}} \right)

, if the eccentricity of the
hyperbola, x²–y²sec²

\theta

= 10 is

\sqrt 5

times the
eccentricity of the ellipse, x²sec²

\theta

+ y² = 5, then the length of the latus rectum of the ellipse, is :

Accepted Answer

Given equation of hyperbola

\Rightarrow {x^2} - {y^2}{\sec ^2}\theta = 10

\Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}\theta }} = 1

Hence eccentricity of hyperbola

\left( {{e_H}} \right) = \sqrt {1 + {{10{{\cos }^2}\theta } \over {10}}}

...(i)

\left\{ { \because \,\, e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} } \right\}

Now equation of ellipse

\Rightarrow {x^2}{\sec ^2}\theta + {y^2} = 5

\Rightarrow {{{x^2}} \over {5{{\cos }^2}}} + {{{y^2}} \over 5} = 1\,

\,\left\{ {e = 1 - {{{a^2}} \over {{b^2}}}} \right\}

Hence eccenticity of ellipse

\left( {{e_E}} \right) = \sqrt {1 - {{5{{\cos }^2}\theta } \over 5}}

\left( {{e_E}} \right) = \sqrt {1 - {{\cos }^2}\theta }

...(ii)

given

{e_H} = \sqrt 5 {e_e}

Hence

\sqrt {1 + {{\cos }^2}\theta } = \sqrt 5 \times \left( {\sqrt {1 - {{\cos }^2}\theta } } \right)

Squaring both sides

1 + {\cos ^2}\theta = 5\left( {1 - {{\cos }^2}\theta } \right)

1 + {\cos ^2}\theta = 5 - 5{\cos ^2}\theta

6{\cos ^2}\theta = 4

{\cos ^2}\theta = {2 \over 3}

...(iii)

Now length of latus rectum of ellipse =

= {{2{a^2}} \over b} = {{10{{\cos }^2}\theta } \over {\sqrt 5 }} = {{20} \over {3\sqrt 5 }} = {{4\sqrt 5 } \over 3}

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