y2=12xa=3SP×SQ=4147 Let P(3t2,6t) and t1t2=−1 (ends of focal chord) So, Q(t23,t−6) $\begin{aligned}
& \mathrm{S}(3,0) \
& \mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_1 \times \mathrm{QM}_2
\end{aligned}(dist.fromdirectrix)\begin{aligned}
& =\left(3+3 \mathrm{t}^2\right)\left(3+\frac{3}{\mathrm{t}^2}\right)=\frac{147}{4} \
& \Rightarrow \frac{\left(1+\mathrm{t}^2\right)^2}{\mathrm{t}^2}=\frac{49}{12} \
& \mathrm{t}^2=\frac{3}{4}, \frac{4}{3} \
& \mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}}
\end{aligned}considering\mathrm{t}=\frac{-\sqrt{3}}{2}\mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right) \text { and } \mathrm{Q}(4,4 \sqrt{3})Hence,diametriccircle:\begin{aligned}
& (x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0 \
& \Rightarrow x^2+y^2-\frac{25}{4} x-\sqrt{3} y-27=0 \
& \Rightarrow \alpha=400, \beta=1728 \
& \beta-\alpha=1328
\end{aligned}$
Let y^2=12 x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that ( SP)( SQ)= 147 4. Let C be the circle described…
JEE Main 2025 — Mathematics Coordinate Geometry
Let y2=12x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ)=4147. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x2+64y2−αx−643y=β, then β−α is equal to ________.
Held on 29 Jan 2025 · Verified 6 Jul 2026.
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