Question

Normals are drawn from the point $P$ with slopes $m_1\cdot m_2=\alpha$. If $P$ lies on the parabola $y^2=4x$ itself, then $\alpha$ is equal to

Answer 1

Let any point on the parabola $y^2=4x$ is $P(h,k)$.
Hence equation of normal on the given point
$\Rightarrow k=mh-2m-m^3$
$\Rightarrow m^3+m(2-h)+k=0$
$\Rightarrow m_1{m}_2{m}_3=-k$
$\Rightarrow m_3=-\frac{k}{\alpha }$
$\Rightarrow {(-\frac{k}{\alpha })}^3-\frac{k}{\alpha }(2-h)+k=0$
$\Rightarrow k^2={\alpha }^{2}h-2{\alpha }^2+{\alpha }^3$
Thus, the locus of $(h,k)$ is
$\Rightarrow y^2={\alpha }^{2}x-2{\alpha }^2+{\alpha }^3$
On comparing it with $y^2=4x$, we get ${\alpha }^2=4$ and $-2{\alpha }^2+{\alpha }^3=0$
$\Rightarrow \alpha =2$