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

<p>Vertices of hyperbola $= (0,\, \pm \,8)$</p> <p>As ellipse pass through it i.e.,</p> <p>$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$ ...... (1)</p> <p>As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.</p> <p>${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}$</p> <p>and ${e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}$</p> <p>$\therefore$ ${e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}$</p> <p>$\Rightarrow (64 - {a^2})(113) = {32^2}$</p> <p>$\Rightarrow {a^2} = 64 - {{1024} \over {113}}$</p> <p>L.R of ellipse $$ = {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)$$</p> <p>$= l = {{1552} \over {113}}$</p> <p>$\therefore$ $113l = 1552$</p>