JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 17)
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 :
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 :
$$\sqrt {30} $$
$$2\sqrt 6 $$
$${{4\sqrt 5 } \over 3}$$
$${{2\sqrt 5 } \over 3}$$
Explanation
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 = $$ = {{2{a^2}} \over b} = {{10{{\cos }^2}\theta } \over {\sqrt 5 }} = {{20} \over {3\sqrt 5 }} = {{4\sqrt 5 } \over 3}$$
$$ \Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}\theta }} = 1$$
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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