JEE Main 2026 (Online) 6th April Morning ShiftQuestion 1
If the eccentricity $e$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, passing through $(6,4\sqrt3)$, satisfies $15(e^{2}+1)=34e$, then the length of the latus rectum of the hyperbola $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{2(a^{2}+1)}=1$ is :
Hyperbola
Solution
$15e^2-34e+15=0$ gives $e=\frac53$, so $b^2=\frac{16}9a^2$; $\frac{36}{a^2}-\frac{48}{b^2}=1$ gives $a^2=9$, $b^2=16$. For $\frac{x^2}{16}-\frac{y^2}{20}=1$, latus rectum $=\frac{2\cdot20}4=10$.