JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Hyperbola

The focus at $(-3,0)$ gives $ae=3$, and the latus rectum through the other focus $(3,0)$ has endpoints $\left(3,\pm \frac{b^2}{a}\right)$. For these endpoints to subtend a right angle at $(-3,0)$ each makes $45^{\circ }$ with the axis, so $\frac{b^2/a}{6}=1$, giving $b^2=6a$. Substituting into $a^2{e}^2=a^2+b^2$ gives $9=a^2+6a$, so $a=3(\sqrt{2}-1)$. Then $a^2{b}^2=6a^3=6(135\sqrt{2}-189)=810\sqrt{2}-1134$, so $\alpha =810$ and $\beta =1134$, and $\alpha +\beta =1944$.