How to find the triangle area inside the parabola? Please help me understand.

Question

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.

Answer 1

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: $$ area_{PSQ}=area_{PRS}+area_{QRS}={1\over2}15\cdot12+{1\over2}15\cdot18=225. $$

Answer 2

The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=\frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by $$y=-px+n$$ and plug in the coordinates of $P$ we get $$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line. And für $p=2$ you have to solve $$(-2x+36)^2=12x$$ for $$x$$