JEE Main 2025 — 29 January, Evening Shift — Question 66

$y^{2}=12 x ,\quad a=3 ,\quad S P \times S Q=\frac{147}{4}$

Let $\mathrm{P}\left(3 \mathrm{t}^{2}, 6 \mathrm{t}\right)$ and $\mathrm{t}_{1} \mathrm{t}_{2}=-1$ (ends of focal chord)

So, $Q\left(\frac{3}{t^{2}}, \frac{-6}{t}\right)$

$\mathrm{S}(3,0)$

$\mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_{1} \times \mathrm{QM}_{2}$

(dist. from directrix)

$=\left(3+3 \mathrm{t}^{2}\right)\left(3+\frac{3}{\mathrm{t}^{2}}\right)=\frac{147}{4}$

$\Rightarrow \frac{\left(1+\mathrm{t}^{2}\right)^{2}}{\mathrm{t}^{2}}=\frac{49}{12}$

$\mathrm{t}^{2}=\frac{3}{4}, \frac{4}{3}$

$\mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}}$

considering $\mathrm{t}=\frac{-\sqrt{3}}{2}$

$\mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right)$ and $\mathrm{Q}(4,4 \sqrt{3})$

Hence, diametric circle: $(x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0$

$\Rightarrow x^{2}+y^{2}-\frac{25}{4} x-\sqrt{3} y-27=0$

$\Rightarrow \alpha=400, \beta=1728$

$\beta-\alpha=1328$