Question

If a variable chord $PQ$ of the parabola $y^2=4ax$ is drawn parallel to $y=x,$ then the locus of point of intersection of normals at $P$ and $Q$ is

• A. $2x-y-12a=0$
• B. $2x-y+10a=0$
• C. $2x-y-8a=0$
• D. $2x-y+6a=0$

Answer 1

The correct option is A $2x-y-12a=0$

Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ are the points where normals are drawn.
Slope of chord $PQ$ is $\frac{2}{t_1+t_2}=1\Rightarrow t_1+t_2=2\cdots \left(1\right)$
Let locus of point of intersection of normals be $R(h,k).$
$(h,k)=\left(a\right(t_1^2+t_2^2+t_1{t}_2+2),-a{t}_1{t}_2(t_1+t_2\left)\right)$
$h=2a+a\left(\right(t_1+t_2{)}^2-t_1{t}_2)$
$\Rightarrow h=2a+a(2^2-t_1{t}_2)\left[\text{From}\right(1\left)\right]$
$\Rightarrow h-6a=-a{t}_1{t}_2$
and $k=-a{t}_1{t}_2(t_1+t_2)=(h-6a)2$
So, required locus is $2x-y-12a=0$