Mathematics - Parabola Question with Solution | TestHub

${\mathrm{y}}^2=12\mathrm{x}\mathrm{a}=3\mathrm{SP}\times \mathrm{SQ}=\frac{147}{4}$ 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, $Q\left(\frac{3}{t^2},\frac{-6}{t}\right)$ $\begin{array}{rl}& \mathrm{S}(3,0)\\ & \mathrm{SP}\times \mathrm{SQ}={\mathrm{PM}}_1\times {\mathrm{QM}}_2\end{array}$ (dist. from directrix) $\begin{array}{rl}& =\left(3+3{\mathrm{t}}^2\right)\left(3+\frac{3}{{\mathrm{t}}^2}\right)=\frac{147}{4}\\ & \Rightarrow \frac{{\left(1+{\mathrm{t}}^2\right)}^2}{{\mathrm{t}}^2}=\frac{49}{12}\\ & {\mathrm{t}}^2=\frac{3}{4},\frac{4}{3}\\ & \mathrm{t}=\pm \frac{\sqrt{3}}{2},\pm \frac{2}{\sqrt{3}}\end{array}$ 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: $\begin{array}{rl}& (x-4)\left(x-\frac{9}{4}\right)+(y+3\sqrt{3})(y-4\sqrt{3})=0\\ & \Rightarrow x^2+y^2-\frac{25}{4}x-\sqrt{3}y-27=0\\ & \Rightarrow \alpha =400,\beta =1728\\ & \beta -\alpha =1328\end{array}$