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0 $begingroup$ The parabola $C$ has cartesian equation $y^2 = 12x.$ The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$ (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. analytic-geometry conic-sections share | cite | improve this question edited Dec 31 '18 at 11:41 ...