JEE Main 2025 — 23 January, Morning Shift — Question 23

$\text{Let\; the\; centre\; of\; }C\; \text{ be\; }(h,0)\ (h>0).$ $\text{Distance\; from\; }(h,0)\text{ to\; }x-y+1=0\text{ is\; } \dfrac{h+1}{\sqrt2},\ \text{so\; } r=\dfrac{h+1}{\sqrt2}.$ $\text{Distance\; from\; }(h,0)\text{ to\; }-3x+2y-1=0\;\text{ is\; } \dfrac{3h+1}{\sqrt{13}}.$ $\text{Chord\; length\; }l=2\sqrt{r^2-\dfrac{(3h+1)^2}{13}}=\dfrac{4}{\sqrt{13}}.$ $\Rightarrow 4\left(r^2-\dfrac{(3h+1)^2}{13}\right)=\dfrac{16}{13} ;\Longrightarrow;13r^2-(3h+1)^2=4.$ $r^2=\dfrac{(h+1)^2}{2} \Rightarrow 13\frac{(h+1)^2}{2}-(3h+1)^2=4.$ $\Rightarrow 5h^2-14h-3=0 \Rightarrow h=\frac{14\pm16}{10} \Rightarrow h=3\ (\text{positive\; root}).$ $\therefore r=\dfrac{3+1}{\sqrt2}=\dfrac{4}{\sqrt2}=2\sqrt2,; r^2=8.$

$\text{Hyperbola\; } \dfrac{x^2}{\alpha^2}-\dfrac{y^2}{\beta^2}=1 \text{ has\; foci\; }(\pm c,0),\ c^2=\alpha^2+\beta^2.$ $\text{One\; focus\; is\; }(3,0)\Rightarrow c=3\Rightarrow \alpha^2+\beta^2=9.$ $\text{Transverse\; axis\; length\; }2\alpha=\text{diameter\; of\; }C=2r=4\sqrt2\Rightarrow \alpha=2\sqrt2\Rightarrow \alpha^2=8.$ $\Rightarrow \beta^2=9-8=1.$

$\therefore 2\alpha^2+3\beta^2=2\cdot8+3\cdot1=16+3=19.$

$\boxed{19}$