Question

A variable chord $PQ$ of the parabola $y^2=4x$ is drawn parallel to the line $y=x$. The locus of point of intersection of normals at $P$ and $Q$ is $\alpha x-y=l$, the value of $\frac{\alpha +l}{2}$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is A $7$

Let $P\equiv (t_1^2,2t_1)$ and $Q\equiv (t_2^2,2t_2)$
Thus the equation of chord $PQ$ is, $y=\frac{2t_1{t}_2}{t_1+t_2}+\frac{2}{t_1+t_2}x$
But given that $PQ$ is parallel to the line $y=x$
$\Rightarrow \frac{2}{t_1+t_2}=1\Rightarrow t_1+t_2=2..\left(1\right)$
Now equation of normal to the parabola $y^2=4x$ at $t$ is given by,
$y+tx=2t+t^3$
Let $P(h,k)$ be the point of intersection of the normals at $P$ and $Q$
$\Rightarrow k+th=2t+t^3$
$\Rightarrow t^3+(2-h)t-k=0$
Clearly this is cubic in $t$ so it will contain three roots
Let $t_1,t_2$ are corresponding to $P,Q$ and $t_3$ is corresponding to some other point.
Thus, $t_1+t_2+t_3=0,t_1{t}_2+t_2{t}_3+t_3{t}_1=2-h$ and $t_1{t}_2{t}_3=k$
Using (1) $t_3=-2$ and $t_1{t}_2=-\frac{k}{2}$
And $t_1{t}_2+t_2{t}_3+t_3{t}_1=2-h\Rightarrow -\frac{k}{2}+t_3(t_1+t_2)=2-h$
$\Rightarrow -\frac{k}{2}-2\times 2=2-h$
$\Rightarrow -\frac{k}{2}-2\times 2=2-h$
$\Rightarrow 2h-k=6$
Hence locus of $P(h,k)$ is, $2x-y=6$
$\Rightarrow \frac{\alpha +l}{2}=\frac{2+12}{2}=7$