Question

PG, the normal at P to a parabola, cuts the axis in G and is produced to Q so that $GQ=\frac{1}{2}PG$; prove that the other normals which pass through Q intersect at right angles.

Answer 1

Let the point $P$ be $(a{t}^2,2at)$

Equation of normal at $P$ is

$y=-tx+2at+a{t}^3$

$a{t}^3+(2a-x)t-y=\mathrm{0.......}\left(i\right)$

It intersect the axis at $G(2a+a{t}^2,0)$

Now $G$ divides $PQ$ in $2:1$

Let the point $q$ be $(h,k)$

$(\frac{2h+a{t}^2}{3},\frac{2k+2at}{3})\equiv (2a+a{t}^2,0)\Rightarrow h=3a+a{t}^2,k=-at$

So the coordinates of $Q$ are $(3a+a{t}^2,-at)$

Now $\left(i\right)$ is cubic in $t$

$\Rightarrow t_1{t}_2{t}_3=\frac{y}{a}$

It passes through $Q$ so $y=-at$ and $t_1=t$

$t{t}_2{t}_3=\frac{-at}{a}{t}_2{t}_3=-1$

So the other two normals are perpendicular