JEE Main 2025 — Mathematics Coordinate Geometry

$\mathrm{y}^2=12 \mathrm{x} \quad \mathrm{a}=3 \quad \mathrm{SP} \times \mathrm{SQ}=\frac{147}{4}$y2=12xa=3SP×SQ=4147​ Let $\mathrm{P}\left(3 \mathrm{t}^2, 6 \mathrm{t}\right)$P(3t2,6t) and $\mathrm{t}_1 \mathrm{t}_2=-1$t1​t2​=−1 (ends of focal chord) So, $Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)$Q(t23​,t−6​) $\begin{aligned}
& \mathrm{S}(3,0) \
& \mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_1 \times \mathrm{QM}_2
\end{aligned}$(dist. from directrix)$(dist.fromdirectrix)\begin{aligned}
& =\left(3+3 \mathrm{t}^2\right)\left(3+\frac{3}{\mathrm{t}^2}\right)=\frac{147}{4} \
& \Rightarrow \frac{\left(1+\mathrm{t}^2\right)^2}{\mathrm{t}^2}=\frac{49}{12} \
& \mathrm{t}^2=\frac{3}{4}, \frac{4}{3} \
& \mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}}
\end{aligned}$considering$considering\mathrm{t}=\frac{-\sqrt{3}}{2}$$\mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right) \text { and } \mathrm{Q}(4,4 \sqrt{3})$Hence, diametric circle:$Hence,diametriccircle:\begin{aligned}
& (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
\end{aligned}$