Let the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is :

<p>Ellipse : ${{{x^2}} \over {16}} + {{{y^2}} \over 7} = 1$</p> <p>Eccentricity $= \sqrt {1 - {7 \over {16}}} = {3 \over 4}$</p> <p>Foci $\equiv ( \pm \,a\,e,0) \equiv ( \pm \,3,0)$</p> <p>Hyperbola : $${{{x^2}} \over {\left( {{{144} \over {25}}} \right)}} - {{{y^2}} \over {\left( {{\alpha \over {25}}} \right)}} = 1$$</p> <p>Eccentricity $= \sqrt {1 + {\alpha \over {144}}} = {1 \over {12}}\sqrt {144 + \alpha }$</p> <p>Foci $$ \equiv ( \pm \,a\,e,0) \equiv \left( { \pm \,{{12} \over 5}\,.\,{1 \over {12}}\sqrt {144 + \alpha } ,\,0} \right)$$</p> <p>If foci coincide then $3 = {1 \over 5}\sqrt {144 + \alpha } \Rightarrow \alpha = 81$</p> <p>Hence, hyperbola is $${{{x^2}} \over {{{\left( {{{12} \over 5}} \right)}^2}}} - {{{y^2}} \over {{{\left( {{9 \over 5}} \right)}^2}}} = 1$$</p> <p>Length of latus rectum $= 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}$</p>