Question

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

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

Answer 1

The correct option is A $343y^2=12(x-6{)}^3$

Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$
As it is given that $2\times 2a{t}_1=2a{t}_2$
$\Rightarrow t_2=2t_1$
Now, point of intersection of normals at $P$ and $Q$ will be
$(2a+a(t_1^2+t_2^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2\left)\right)$
But here $a=3$ and $t_2=2t_1$
Hence, point of intersection will be
$\left(3\right(2+7t_1^2),-18t_1^3)\equiv (h,k)$
$t_1=-\sqrt[3]{3\frac{k}{18}}$
$h=6+21{(-\sqrt[3]{3\frac{k}{18}})}^2\Rightarrow 343k^2=12(h-6{)}^3$
Hence, locus will be $343y^2=12(x-6{)}^3$

Alternate Solution :
Let $(h,k)$ is the point of intersection of the normals.
So, normal equation passing through $(h,k)$ is
$\therefore k+th=2at+a{t}^3...\left(i\right)$
$\Rightarrow a{t}^3+(2a-h)t-k=0$
$t_1{t}_2{t}_3=\frac{k}{a},t_1+t_2+t_3=0$
Also given, $\frac{2a{t}_1}{2a{t}_2}=\frac{1}{2}$
$\Rightarrow 2t_1=t_2\Rightarrow t_3=-3t_1$
$\therefore \left(t_1\right)\left(2t_1\right)(-3t_1)=\frac{k}{a}$
$\Rightarrow t_1^3=-\frac{k}{6a}$
Since, $t_1$ satisfies equation $\left(i\right)$
$\therefore (2a-h{)}^3{t}_1^3=(k-a{t}_1^3{)}^3$
$\Rightarrow -(2a-h{)}^3\frac{k}{6a}={(k+\frac{k}{6})}^3$
$\therefore -(6-x{)}^3=\frac{343}{12}{y}^2$
i.e., $343y^2=-12(6-x{)}^3=12(x-6{)}^3$