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

Answer: 3

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