If the ordinates of points $P$ and $Q$ on the parabola $y^2=12 x$ are in the ratio $1: 2$, then the locus of…

If the ordinates of points $P$ and $Q$ on the parabola $y^2=12 x$ are in the ratio $1: 2$, then the locus of the point of intersection of the normals to the parabola at P and Q is
  1. $y+18\left(\frac{x-6}{21}\right)^{3 / 2}=0$
  2. $y-18\left(\frac{x-6}{12}\right)^{3 / 2}=0$
  3. $y+12\left(\frac{x-6}{14}\right)^{1 / 2}=0$
  4. $y-12\left(\frac{x-6}{18}\right)^{3 / 2}=0$

Solution

Given, $\frac{t_1}{t_2}=2$
Let $t_2=t \Rightarrow t_1=2 t$ Point of intersection of normals at $t_1$ and $t_2$ is $\begin{aligned} & \left(2 a+a\left(t_1^2+t_2^2+t_1 t_2\right),-a t_1 t_2\left(t_1+t_2\right)\right. \\ & \left(2 a+a\left(4 t^2+t^2+2 t^2\right),-2 a t^2 \times 3 t\right) \equiv\left(2 a+7 a t^2,-6 a t^3\right) \\ & x=2 a+7 a t^2, y=-6 a t^3 \\ & \Rightarrow x-2 a=7 a t^2, y=-6 a t^3 \\ & \Rightarrow a=3 \Rightarrow x-6=21 t^2, y=-18 t^3 \end{aligned}$
Eliminating $t$ we get, $y+18\left(\frac{x-6}{21}\right)^{3 / 2}=0$.

Asked in: AP EAMCET 2024 (21 May Shift 2)

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