Question

$P$ is a point on the parabola $y^{2}=12x$ such that $PS=6$ ( $S$ is the focus of the parabola). Tangent is drawn at $P$ to parabola which intersects the tangent at vertex at $T$ . A point $R$ is taken on the axis of parabola such that $SR=6$ and $R$ lies inside the parabola. The area of quadrilateral $PRST$ is $A$ then the value of $\frac{A}{9}$ is

Answer 1

Area of quadrilateral $PRST$
$=$ Area of $\Delta PQR-$ Area of $\Delta QTS$
$6.y=6\left(x + 3\right)$
$T\left(0 , 3\right)$ $x-y+3=0$ $Q\left(- 3 , 0\right) \, $
$y-6=-1\left(x - 3\right)\Rightarrow y-6=-x+3$
$x+4=9$
$R\left(9 , 0\right)$
$PR=\sqrt{\left(3 + 3\right)^{2} + \left(6 - 0\right)^{2}}=\sqrt{36 + 36}=\sqrt{72}$
$=6\sqrt{2}$
$PR=6\sqrt{2},$ $TS=QT$
$=\sqrt{9 + 9}=3\sqrt{2}$
$A=\frac{1}{2}\times 6\sqrt{2}\times 6\sqrt{2}-\frac{1}{2}\times 3\sqrt{2}\times 3\sqrt{2}$
$\Rightarrow \frac{1}{2}\left(72 - 18\right)^{2}=\frac{54}{2}=27$