Question

The ordinates of points $P$ and $Q$ on the parabola $y^2=12x$ are in the ratio $1:2$ . Find the locus of the point of intersection of the normals to the parabola at $P$ and $Q$.

Answer 1

We have $y^2=12x$

$\Rightarrow 4a=12$

$\Rightarrow a=\frac{12}{4}=3$

Let co−ordinates of $P$ be $(3{t_1}^2,6t_1)$ and that of $Q$ be $(3{t_2}^2,6t_2)$

Given that $\frac{6t_2}{6t_1}=2$

$\Rightarrow t_2=2t_1$ .......$\left(1\right)$

Let $R(h,k)$ be point of intersection of normal.

We know that point of intersection of normal at $t_1$ and $t_2$ to parabola $y^2=4ax$ is

$(a({t_1}^2+{t_2}^2+t_1{t}_2+2),-a{t}_1{t}_2(t_1+t_2))$

$\Rightarrow h=3({t_1}^2+{t_2}^2+t_1{t}_2+2)$

$\Rightarrow h=3{t_1}^2+3{t_2}^2+3t_1{t}_2+6$

$\Rightarrow h=3{t_1}^2+3{\left(2t_1\right)}^2+3t_1\left(2t_1\right)+6$

$\Rightarrow h=3{t_1}^2+12{t_1}^2+6{t_1}^2+6$

$\Rightarrow h=21{t_1}^2+6$

$\Rightarrow {t_1}^2=\frac{h-6}{21}$ .......$\left(2\right)$

$k=-a{t}_1{t}_2(t_1+t_2)$

$\Rightarrow k=-3t_1\times 2t_1(t_1+2t_1)$

$\Rightarrow k=-6{t_1}^2\left(3t_1\right)$

$\Rightarrow k=-18{t_1}^3$

$\Rightarrow k^2=324{\left({t_1}^3\right)}^2$

$\Rightarrow k^2=324{\left({t_1}^2\right)}^3$

$\Rightarrow k^2=324{\left(\frac{h-6}{21}\right)}^3$

$\Rightarrow k^2=\frac{324}{21\times 21\times 21}{(h-6)}^3$

$\Rightarrow k^2=\frac{12}{343}{(h-6)}^3$

$\Rightarrow 343k^2=12{(h-6)}^3$

Replace $h\to x$ and $k\to y$ we get

$343y^2=12{(x-6)}^3$ is the locus of the point of intersection of the normals to the parabola at $P$ and $Q$.