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

Answer 1

Answer: (A)

Here, $a=3.$ So, let $P\left(3 t_{1}^{2} , 6 t_{1}\right)$ and $Q\left(3 t_{2}^{2} , 6 t_{2}\right)$ be two points on the parabola $y^{2}=12x.$
Then,
$\frac{6 t_{1}}{6 t_{2}}=\frac{1}{2}\Rightarrow t_{2}=2t_{1}.$
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 $k=-3t_{1}t_{2}\left(t_{1} + t_{2}\right)$
$\Rightarrow h=6+3\left(t_{1}^{2} + 4 t_{1}^{2} + 2 t_{1}^{2}\right)$ and $k=-6t_{1}^{2}\left(3 t_{1}\right)$
$\Rightarrow h-6=21t_{1}^{2}$ 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
$y^{2}=\frac{12}{343}\left(x - 6\right)^{3}$