Let the foci of the ellipse x ^ 2 16 + y ^ 2 7 = 1 and the...

Given equation of ellipse $\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{7} = 1$

Now finding eccentricity $= \sqrt{1 - \dfrac{7}{16}} = \dfrac{3}{4}$

So, foci $≡ \left(\pm ae, 0\right) ≡ \left(\pm 3, 0\right)$

Now, hyperbola: $\dfrac{{x}^{2}}{\left(\dfrac{144}{25}\right)} - \dfrac{{y}^{2}}{\left(\dfrac{\alpha}{25}\right)} = 1$

Eccentricity will be $= \sqrt{1 + \dfrac{\alpha}{144}} = \dfrac{1}{12}\sqrt{144 + \alpha}$

Foci $≡ \left(\pm ae, 0\right) ≡ \left(\pm \dfrac{12}{5} \cdot \dfrac{1}{12}\sqrt{144 + \alpha}, 0\right)$

Given foci coincide then $3 = \dfrac{1}{5}\sqrt{144 + \alpha} ⇒ \alpha = 81$

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 $= 2\dfrac{{b}^{2}}{a} = 2 \cdot \dfrac{\dfrac{81}{25}}{\dfrac{12}{5}} = \dfrac{27}{10}$