Let the foci of the ellipse x^216+y^27=1 and the hyperbola x^2144-y^2=125 coincide.

Question

Let the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}7=1$ and the hyperbola $\dfrac{x^2}{144}-\dfrac{y^2}\alpha=\dfrac1{25}$ coincide. Then the length of the latus rectum of the hyperbola is:

Accepted Answer

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 $$ = 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}$$

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