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Q.
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:-
$\frac{x^2}{16}+\frac{y^2}{7}=1 $
$\Rightarrow 7=16\left(1- e ^2\right) \Rightarrow e =\frac{3}{4}$
Foci of ellipse is $(\pm a e, 0) \Rightarrow(\pm 3,0)$
Now hyperbola be $\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}$
$\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{\alpha}{25}}=1$
Now $a =\frac{12}{5}, b ^2=\frac{\alpha}{25}$
Let eccentricity of hyperbola be e ae $=3$ (Given)
$ \Rightarrow \frac{12}{5} e =3 \Rightarrow e =\frac{5}{4} $
$b ^2= a ^2\left( e ^2-1\right)$
$\frac{\alpha}{25}=\frac{144}{25}\left(\frac{25}{16}-1\right) \Rightarrow \alpha=81$
Hyperbola is $\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$
Now length of $L R=\frac{2 b^2}{a}=\frac{27}{10}$