JEE Main 2026 — Hyperbola Question with Solution

The given equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Since it passes through $(6,4\sqrt{3})$, we have:

$\frac{36}{a^2}-\frac{48}{b^2}=1$

The eccentricity $e$ satisfies $15(e^2+1)=34e$.

$15e^2-34e+15=0$

$(3e-5)(5e-3)=0$

Since the eccentricity of a hyperbola is $e>1$, we get $e=\frac{5}{3}$.

We know that $b^2=a^2(e^2-1)$.

$b^2=a^2\left(\frac{25}{9}-1\right)=\frac{16}{9}{a}^2$

Substituting $b^2$ into the first equation:

$\frac{36}{a^2}-\frac{48}{\frac{16}{9}{a}^2}=1$

$\frac{36}{a^2}-\frac{27}{a^2}=1$

$\frac{9}{a^2}=1\Rightarrow a^2=9$

Then, $b^2=\frac{16}{9}\times 9=16$.

The second hyperbola is given by $\frac{x^2}{b^2}-\frac{y^2}{2(a^2+1)}=1$.

Substituting the values of $a^2$ and $b^2$:

$\frac{x^2}{16}-\frac{y^2}{2(9+1)}=1$

$\frac{x^2}{16}-\frac{y^2}{20}=1$

Here, $A^2=16\Rightarrow A=4$ and $B^2=20$.

The length of the latus rectum is $\frac{2B^2}{A}$.

Length of latus rectum $=\frac{2\times 20}{4}=10$.

Answer: $10$