Let the latus rectum of the hyperbola $\frac{x^2}{9}-\; <!--\; -\; -->\frac{y^2}{b^2}=1$ subtend an angle of $\frac{\mathrm{\pi \; <!--\; \pi \; -->}}{3}$ at the centre of the hyperbola. If b² is equal to $\frac{\mathrm{l}}{\mathrm{m}}(1+\sqrt{\mathrm{n}}),$where $\mathrm{l}$ and m are co-prime numbers, then ${\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2$ is equal to ____

The correct answer is 182.

We are given the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$, so $a^2=9$, meaning $a=3$.

Step 1: Identify the endpoints of the latus rectum.

For a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the latus rectum passes through the focus $(ae,0)$. Its two endpoints are:

$(ae,\frac{b^2}{a})$ and $(ae,-\frac{b^2}{a})$

Step 2: Set up the angle condition.

The latus rectum subtends an angle of $\frac{\pi }{3}$ at the centre (origin). By symmetry, each endpoint makes a half-angle of $\frac{\pi }{6}$ with the x-axis. Therefore:

$\tan \left(\frac{\pi }{6}\right)=\frac{b^2/a}{ae}=\frac{b^2}{a^{2}e}$

Since $\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$, we get:

$\frac{b^2}{a^{2}e}=\frac{1}{\sqrt{3}}⟹b^2=\frac{a^2}{e}=\frac{9}{e}=3\sqrt{3}e$

Step 3: Use the hyperbola relation $b^2=a^2(e^2-1)$.

$9(e^2-1)=3\sqrt{3}e$

Dividing through by 9:

$e^2-\frac{e}{\sqrt{3}}-1=0$

Step 4: Solve the quadratic for $e$.

Multiplying by $\sqrt{3}$ to clear:

$\sqrt{3}{e}^2-e-\sqrt{3}=0$

Using the quadratic formula with coefficients $A=\sqrt{3}$, $B=-1$, $C=-\sqrt{3}$:

$e=\frac{1\pm \sqrt{1+4·3}}{2\sqrt{3}}=\frac{1+\sqrt{13}}{2\sqrt{3}}$ (taking the positive root, since $e>1$)

Step 5: Find $b^2$.

First, compute $e^2$:

$e^2=\frac{(1+\sqrt{13}){}^2}{4·3}=\frac{1+2\sqrt{13}+13}{12}=\frac{14+2\sqrt{13}}{12}=\frac{7+\sqrt{13}}{6}$

Then:

$e^2-1=\frac{7+\sqrt{13}}{6}-1=\frac{1+\sqrt{13}}{6}$

So:

$b^2=a^2(e^2-1)=9·\frac{1+\sqrt{13}}{6}=\frac{3}{2}(1+\sqrt{13})$

Step 6: Identify l, m, and n.

We have $b^2=\frac{l}{m}(1+\sqrt{n})$, which gives us:

l = 3, m = 2, n = 13

We verify that l and m are co-prime: gcd(3, 2) = 1 ✓

Step 7: Compute the final answer.

$l^2+m^2+n^2=3^2+2^2+{13}^2=9+4+169=$182