Question

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

• A. $y^2=\frac{12}{343}{\left(x-6\right)}^3$
• B. $y^2=\frac{12}{343}{\left(x+6\right)}^3$
• C. ${\left(y-6\right)}^2=\frac{12x}{343}$
• D. ${\left(y+6\right)}^2=\frac{12x}{343}$

Answer 1

Answer: (A)

Here, $a=3.$ So, let $P\left(3t_1^2,6t_1\right)$ and $Q\left(3t_2^2,6t_2\right)$ be two points on the parabola $y^2=12x.$
Then,
$\frac{6t_1}{6t_2}=\frac{1}{2}\Rightarrow t_2=2t_1.$
Let, $\left(h,k\right)$ be the point of intersection of the normals at $P$ and $Q.$
Then,
$k=-t_{1}h+6t_1+3t_1^3$ and $k=-t_{2}h+6t_2+3t_2^3$
$\Rightarrow h=6+3\left(t_1^2+t_2^2+t_1{t}_2\right)$ and $k=-3t_1{t}_2\left(t_1+t_2\right)$
$\Rightarrow h=6+3\left(t_1^2+4t_1^2+2t_1^2\right)$ and $k=-6t_1^2\left(3t_1\right)$
$\Rightarrow h-6=21t_1^2$ and $k=-18t_1^3$
$\Rightarrow {\left(\frac{h-6}{21}\right)}^3=t_1^6$ and ${\left(\frac{-k}{18}\right)}^2=t_1^6$
$\Rightarrow {\left(\frac{h-6}{21}\right)}^3=\frac{k^2}{324}$
$\Rightarrow k^2=\frac{12}{343}{\left(h-6\right)}^3$
Hence, the locus of $\left(h,k\right)$ is
$y^2=\frac{12}{343}{\left(x-6\right)}^3$