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QJEEM22Ma9f7d: JEE Main 2022 Mathematics - Class 11 JEEM22Ma9f7d Solution: An ellipse passes through the vertices of the hyperbola . Let…

An ellipse E:x2a2+y2b2=1E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 passes through the vertices of the hyperbola H:x249−y264=−1H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1 . Let the major and minor axes of the ellipse EE coincide with the transverse and conjugate axes of the hyperbola HH , respectively. Let the product of the eccentricities of EE and HH be 12\frac{1}{2} . If ll is the length of the latus rectum of the ellipse EE , then the value of 113l113 l is equal to _____________.

Step-by-Step Verified Solution

Vertices of hyperbola =(0, ± 8)= (0,\, \pm \,8)

As ellipse pass through it i.e.,

0+64b2=1⇒b2=640 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64 ...... (1)

As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.

eE=1−a264=64−a28{e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}

and eH=1+4964=1138{e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}

∴\therefore eE . eH=12=64−a211364{e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}

⇒(64−a2)(113)=322\Rightarrow (64 - {a^2})(113) = {32^2}

⇒a2=64−1024113\Rightarrow {a^2} = 64 - {{1024} \over {113}}

L.R of ellipse =2a2b=28(113×64−1024113)= {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)

=l=1552113= l = {{1552} \over {113}}

∴\therefore 113l=1552113l = 1552

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