Let y^2=12 x be the parabola and S be its focus.

Question

Let $y^2=12 x$ be the parabola and $S$ be its focus. Let $P Q$ be a focal chord of the parabola such that $(S P)(S Q)=\frac{147}{4}$. Let $C$ be the circle described taking $P Q$ as a diameter. If the equation of a circle $C$

Accepted Answer

${y}^{2}=12x\, a=3\, SP	imes SQ=\frac{147}{4}$

Let $P(3{t}^{2},6t)$ and ${t}_{1}{t}_{2}=-1$

(ends of focal chord)

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

$S(3,0)$

$SP	imes SQ={PM}_{1}	imes {QM}_{2}$

$\begin{gathered} =(3+3{t}^{2})(3+\frac{3}{{t}^{2}})=\frac{147}{4} \ \Rightarrow \frac{{(1+{t}^{2})}^{2}}{{t}^{2}}=\frac{49}{12} \ {t}^{2}=\frac{3}{4},\frac{4}{3} \ t=\pm \frac{\sqrt{3}}{2},\pm \frac{2}{\sqrt{3}} \ t=\frac{-\sqrt{3}}{2} \ P(\frac{9}{4},-3\sqrt{3})Q(4,4\sqrt{3}) \end{gathered}$

$\begin{gathered} \ (x-4)(x-\frac{9}{4})+(y+3\sqrt{3})(y-4\sqrt{3})=0 \ \Rightarrow {x}^{2}+{y}^{2}-\frac{25}{4}x-\sqrt{3}y-27=0 \ \Rightarrow \alpha =400,\beta =1728 \ \beta -\alpha =1328 \end{gathered}$

Solutions · 1