QJEEM22Ma0bbf: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0bbf Solution: Let the foci of the ellipse and the hyperbola coincide. Then…

Ellipse : ${{{x^2}} \over {16}} + {{{y^2}} \over 7} = 1$

Eccentricity $= \sqrt {1 - {7 \over {16}}} = {3 \over 4}$

Foci $\equiv ( \pm \,a\,e,0) \equiv ( \pm \,3,0)$

Hyperbola : ${{{x^2}} \over {\left( {{{144} \over {25}}} \right)}} - {{{y^2}} \over {\left( {{\alpha \over {25}}} \right)}} = 1$

Eccentricity $= \sqrt {1 + {\alpha \over {144}}} = {1 \over {12}}\sqrt {144 + \alpha }$

Foci $\equiv ( \pm \,a\,e,0) \equiv \left( { \pm \,{{12} \over 5}\,.\,{1 \over {12}}\sqrt {144 + \alpha } ,\,0} \right)$

If foci coincide then $3 = {1 \over 5}\sqrt {144 + \alpha } \Rightarrow \alpha = 81$

Hence, hyperbola is ${{{x^2}} \over {{{\left( {{{12} \over 5}} \right)}^2}}} - {{{y^2}} \over {{{\left( {{9 \over 5}} \right)}^2}}} = 1$

Length of latus rectum $= 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}$