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

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