JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Hyperbola

The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has eccentricity satisfying $15e^2-34e+15=0$, which factors as $(3e-5)(5e-3)=0$, so $e=\frac{5}{3}$ or $e=\frac{3}{5}$; a hyperbola has $e>1$, hence $e=\frac{5}{3}$. From $e^2=1+\frac{b^2}{a^2}$, $\frac{b^2}{a^2}=\frac{25}{9}-1=\frac{16}{9}$. The point $(6,4\sqrt{3})$ lies on the curve, so $\frac{36}{a^2}-\frac{48}{b^2}=1$; putting $b^2=\frac{16}{9}{a}^2$ gives $\frac{36}{a^2}-\frac{27}{a^2}=1$, so $a^2=9$ and $b^2=16$. For $\frac{x^2}{b^2}-\frac{y^2}{2(a^2+1)}=1$ the transverse semi-axis is $\sqrt{b^2}=4$ and $2(a^2+1)=20$, so the length of the latus rectum is $\frac{2\cdot 20}{4}=10$.