JEE Mathematics 2024 Question with Solution

Given: The hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$, so $a^2=9$ and hence $a=3$.

Find: The value of $l^2+m^2+n^2$ when $b^2=\frac{l}{m}(1+\sqrt{n})$.

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the focus is at $(c,0)$ where

So here,

The endpoints of the latus rectum through the focus $(c,0)$ are

Since $a=3$,

The latus rectum subtends an angle of $\frac{\pi}{3}$ at the center, so the x-axis bisects this angle. Therefore,

where $M(c,0)$.

In right triangle $OML$,

Thus,

Using $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$,

Rearranging,

Squaring both sides,

Let $Y=b^2$. Then,

Using the quadratic formula,

Since $b^2>0$,

Comparing with $b^2=\frac{l}{m}(1+\sqrt{n})$, we get

Therefore,

Therefore, the required value is $182$.

Given: $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ and the latus rectum subtends $60^\circ$ at the center.

Find: $l^2+m^2+n^2$.

Let the eccentricity be $e$. For the hyperbola, $c=ae$ and the endpoint of the latus rectum is $\left(ae,\frac{b^2}{a}\right)$.

Since the latus rectum subtends $60^\circ$ at the center, the half-angle is $30^\circ$. Hence,

With $a=3$,

Also, for a hyperbola,

Using the relation from the angle condition and simplifying leads to the same quadratic in $b^2$:

So,

Hence $l=3, m=2, n=13$, and

The correct numerical answer is $182$.