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:

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

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

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

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

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

Foci $\equiv \left(\pm ae,0\right)\equiv \left(\pm \frac{12}{5}·\frac{1}{12}\sqrt{144+\alpha },0\right)$

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

Hence, hyperbola is $\frac{x^2}{{\left(\frac{12}{5}\right)}^2}-\frac{y^2}{{\left(\frac{9}{5}\right)}^2}=1$

Length of latus rectum $=2\frac{b^2}{a}=2·\frac{{\displaystyle \frac{81}{25}}}{{\displaystyle \frac{12}{5}}}=\frac{27}{10}$