JEE Main 2026 Jan 28 Shift 1, Mathematics Q22: Introduction to Hyperbola

Hyperbola $\frac{x^2}{8}-\frac{y^2}{8{\cos }^2\theta }=1$: $a^2=8$, $b^2=8{\cos }^2\theta$, so $e_1=\sqrt{1+{\cos }^2\theta }$ and ${\ell }_1=\frac{2b^2}{a}=\frac{16{\cos }^2\theta }{2\sqrt{2}}=\frac{8{\cos }^2\theta }{\sqrt{2}}$.

Ellipse $\frac{x^2}{6{\cos }^2\theta }+\frac{y^2}{6}=1$: $a^2=6$, $b^2=6{\cos }^2\theta$ (with $a>b$ since ${\sec }^2\theta >1$). So $e_2=\sqrt{1-{\cos }^2\theta }=\sin \theta$ and ${\ell }_2=\frac{2b^2}{a}=\frac{12{\cos }^2\theta }{\sqrt{6}}$.

Apply the condition $e_1^2=e_2^2({\sec }^2\theta +1)$:

$1+{\cos }^2\theta ={\sin }^2\theta \left(1+{\sec }^2\theta \right)={\sin }^2\theta +{\tan }^2\theta$.

Hence $1+{\cos }^2\theta ={\sin }^2\theta +{\tan }^2\theta$. Using ${\sin }^2\theta =1-{\cos }^2\theta$: $1+{\cos }^2\theta =1-{\cos }^2\theta +{\tan }^2\theta \Rightarrow {\tan }^2\theta =2{\cos }^2\theta$. This gives ${\sin }^2\theta =2{\cos }^4\theta$; solving yields $\theta =\frac{\pi }{4}$.

At $\theta =\pi /4$: ${\ell }_1=\frac{8\cdot 1/2}{\sqrt{2}}=2\sqrt{2}$; $e_1=\sqrt{3/2}$; ${\ell }_2=\frac{12\cdot 1/2}{\sqrt{6}}=\sqrt{6}$; $e_2=\frac{1}{\sqrt{2}}$; ${\tan }^2\theta =1$.

$\frac{{\ell }_1{\ell }_2}{e_1{e}_2}{\tan }^2\theta =\frac{2\sqrt{2}\cdot \sqrt{6}}{\sqrt{3/2}\cdot 1/\sqrt{2}}\cdot 1=\frac{2\sqrt{12}}{\sqrt{3/4}}=\frac{4\sqrt{3}}{\sqrt{3}/2}=8$.

Answer: $8$.