1. Write the standard facts for each hyperbola

For a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ we use:

• Eccentricity: $e^2=1+\frac{b^2}{a^2}$
• Length of latus rectum: $L=\frac{2b^2}{a}$
• Length of transverse axis: $2a$

For a hyperbola of the form $\frac{y^2}{B^2}-\frac{x^2}{A^2}=1,$ its transverse axis is along the $y$-axis, and similarly:

• $e^2=1+\frac{A^2}{B^2}$
• Length of latus rectum: $L=\frac{2A^2}{B}$
• Length of transverse axis: $2B$

Here, $H_1:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $H_2:-\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\iff \frac{y^2}{B^2}-\frac{x^2}{A^2}=1.$

2. Use the data for $H_1$

Given:

• latus rectum of $H_1$ is $15\sqrt2$
• eccentricity $e_1=\sqrt{\frac52}$

Since $e_1^2=1+\frac{b^2}{a^2},$ we get $\frac52=1+\frac{b^2}{a^2}$ $\Rightarrow \frac{b^2}{a^2}=\frac32$ $\Rightarrow b^2=\frac32 a^2.$

Now use latus rectum: $\frac{2b^2}{a}=15\sqrt2.$ Substitute $b^2=\frac32 a^2$: $\frac{2\cdot \frac32 a^2}{a}=15\sqrt2$ $3a=15\sqrt2$ $a=5\sqrt2.$

Hence transverse axis length of $H_1$ is $2a=10\sqrt2.$

3. Use the product of transverse axes

Given product of lengths of transverse axes is $100\sqrt{10}.$

For $H_1$, transverse axis length is $10\sqrt2$. For $H_2$, transverse axis length is $2B$.

So, $(10\sqrt2)(2B)=100\sqrt{10}$ $20\sqrt2\,B=100\sqrt{10}$ $B=\frac{100\sqrt{10}}{20\sqrt2}=5\sqrt5.$

Thus, $B=5\sqrt5.$

4. Use latus rectum for $H_2$

Given latus rectum of $H_2$ is $12\sqrt5$. For $\frac{y^2}{B^2}-\frac{x^2}{A^2}=1,$ length of latus rectum is $\frac{2A^2}{B}.$

Hence, $\frac{2A^2}{B}=12\sqrt5.$ Substitute $B=5\sqrt5$: $\frac{2A^2}{5\sqrt5}=12\sqrt5$ $2A^2=12\sqrt5\cdot 5\sqrt5=12\cdot 25=300$ $A^2=150.$

Also, $B^2=(5\sqrt5)^2=125.$

5. Find $e_2$

For $H_2$, $e_2^2=1+\frac{A^2}{B^2}.$ So, $e_2^2=1+\frac{150}{125}=1+\frac65=\frac{11}{5}.$

Therefore, $25e_2^2=25\cdot \frac{11}{5}=55.$

6. Final answer

$\boxed{55}$

This matches the stored correct answer.