How to find the triangle area inside the parabola? Please help me understand. The parabola $C$ has cartesian equation $y^2 = 12x.$
The point $P(3p^2, 6p)$ lies on $C,$ where $p\neq0.$


*

*(a) Show that the equation of the normal to the curve $C$ at the point $P$ is
$$y + px = 6p + 3p^3$$
This normal crosses the curve $C$ again at the point $Q.$
Given that $p = 2$ and that $S$ is the focus of the parabola, ﬁnd

*(b) the coordinates of the point $Q,$

*(c) the area of the triangle $PQS.$
I can't figure out a way to solve question (c). I know the answer but don't understand it.

For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1\over2}15\cdot12+{1\over2}15\cdot18=225.
$$
