JEE Main 2025 Jan 24 Shift 2, Mathematics Q24: Introduction to Hyperbola

Hyperbola $H_1$ (transverse axis along $x$):

Latus rectum $=\frac{2b^2}{a}=15\sqrt{2}\dots (\text{i})$

$e_1^2=1+\frac{b^2}{a^2}=\frac{5}{2}\Rightarrow \frac{b^2}{a^2}=\frac{3}{2}\Rightarrow b^2=\frac{3a^2}{2}$

Substitute into (i): $\frac{2\cdot 3a^2/2}{a}=3a=15\sqrt{2}\Rightarrow a=5\sqrt{2}$, hence $b=5\sqrt{3}$.

Hyperbola $H_2$ (transverse axis along $y$, length $=2B$):

Latus rectum $=\frac{2A^2}{B}=12\sqrt{5}\dots (\text{ii})$

Product-of-axes condition: transverse axis of $H_1$ is $2a=10\sqrt{2}$; that of $H_2$ is $2B$. So

$(2a)(2B)=100\sqrt{10}\Rightarrow 10\sqrt{2}\cdot 2B=100\sqrt{10}\Rightarrow B=\frac{100\sqrt{10}}{20\sqrt{2}}=5\sqrt{5}$

From (ii): $A^2=\frac{12\sqrt{5}\cdot B}{2}=6\sqrt{5}\cdot 5\sqrt{5}=150\Rightarrow A=5\sqrt{6}$.

For a hyperbola with vertical transverse axis, $e_2^2=1+\frac{A^2}{B^2}=1+\frac{150}{125}=1+\frac{6}{5}=\frac{11}{5}$.

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

Answer: 55