JEE Main 2025 Apr 7 Shift 1, Mathematics Q23: Introduction to Hyperbola

Set up. Foci: $F_1=(-ae,0)=(-3,0)$ so $ae=3$. Other focus $F_2=(3,0)$. The latus-rectum endpoints through $F_2$ are $\left(3,\pm \frac{b^2}{a}\right)$.

Right-angle condition at $P=F_1$. The two latus-rectum endpoints subtend a right angle at $F_1$. The lines from $F_1=(-3,0)$ to each endpoint have slopes $\pm \frac{b^2/a}{6}$. For a right angle:

$\tan 45^{\circ }=\frac{b^2/a}{2ae}=\frac{b^2}{2a\cdot ae}\Rightarrow 1=\frac{b^2}{2ae\cdot a}\Rightarrow 2ae=\frac{b^2}{a}$

Substituting $ae=3$: $\frac{b^2}{a}=6\Rightarrow b^2=6a$.

Use the hyperbola identity $a^2{e}^2=a^2+b^2$.

$9=a^2+6a\Rightarrow a^2+6a-9=0$

$a=\frac{-6+\sqrt{36+36}}{2}=-3+3\sqrt{2}$ (taking the positive root)

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

Compute $a^2{b}^2$.

$a^2{b}^2=a^2\cdot 6a=6a^3$

$a^3=27(\sqrt{2}-1{)}^3=27(2\sqrt{2}-6+3\sqrt{2}-1)=27(5\sqrt{2}-7)$

$a^2{b}^2=6\cdot 27(5\sqrt{2}-7)=162(5\sqrt{2}-7)=810\sqrt{2}-1134$

Matching with $\alpha \sqrt{2}-\beta$: $\alpha =810$, $\beta =1134$.

$\alpha +\beta =810+1134=\boxed{1944}$