For some heta \in \left( {0,{\pi \over 2}} ight), if the eccentricity of the hyperbola, x2–y2sec2 heta = 10 is \sqrt 5 times the eccentricity of the ellipse, x2sec2 heta + y2 = 5, then the length of the latus rectum of the ellipse, is :

Question

For some 	heta \in \left( {0,{\pi \over 2}} ight), if the eccentricity of the hyperbola, x2–y2sec2	heta = 10 is \sqrt 5 times the eccentricity of the ellipse, x2sec2	heta + y2 = 5, then the length of the latus rectum of the ellipse, is :

Accepted Answer

Answer: C. {{4\sqrt 5 } \over 3} Given equation of hyperbola \Rightarrow {x^2} - {y^2}{\sec ^2}	heta = 10 \Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}	heta }} = 1 Hence eccentricity of hyperbola \left( {{e_H}} ight) = \sqrt {1 + {{10{{\cos }^2}	heta } \over {10}}} ...(i) \left\{ { \because \,\, e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} } ight\} Now equation of ellipse \Rightarrow {x^2}{\sec ^2}	heta + {y^2} = 5 \Rightarrow {{{x^2}} \over {5{{\cos }^2}}} + {{{y^2}} \over 5} = 1\, \,\left\{ {e = 1 - {{{a^2}} \over {{b^2}}}} ight\} Hence eccenticity of ellipse \left( {{e_E}} ight) = \sqrt {1 - {{5{{\cos }^2}	heta } \over 5}} \left( {{e_E}} ight) = \sqrt {1 - {{\cos }^2}	heta } ...(ii) given {e_H} = \sqrt 5 {e_e} Hence \sqrt {1 + {{\cos }^2}	heta } = \sqrt 5 	imes \left( {\sqrt {1 - {{\cos }^2}	heta } } ight) Squaring both sides 1 + {\cos ^2}	heta = 5\left( {1 - {{\cos }^2}	heta } ight) 1 + {\cos ^2}	heta = 5 - 5{\cos ^2}	heta 6{\cos ^2}	heta = 4 {\cos ^2}	heta = {2 \over 3} ...(iii) Now length of latus rectum of ellipse = = {{2{a^2}} \over b} = {{10{{\cos }^2}	heta } \over {\sqrt 5 }} = {{20} \over {3\sqrt 5 }} = {{4\sqrt 5 } \over 3}

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 :

Answer

Explanation

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