Three normals with slopes  $m_1,m_2\text{\hspace{0.17em}and\hspace{0.17em}}{m}_3$  are drawn from a point P not on the axis of the parabola  $y^2=4x.$ If  $m_1{m}_2=\alpha ,$ results in the locus of P being a part of the parabola, find the value of  $\alpha$.

Any normal of slope m to the parabola 
   $y^2=4x\text{\hspace{0.17em}is}y=mx-2m-m^3\left(1\right)$
If it passes through (h, k), then  $k=mh-2m-m^3$
 $\Rightarrow m^3+(2-h)m+k=0\left(2\right)$
Thus  $m_1{m}_2{m}_3=-k$
Now  $m_1{m}_2=\alpha \left(\text{given}\right)\Rightarrow m_3=-\frac{k}{\alpha }$
Now  $m_3$ satisfies equation (2) so  $-\frac{k^3}{{\alpha }^3}-(2-h)\frac{k}{\alpha }+k=0$
$\Rightarrow k^3+(2-h)k{\alpha }^2-k{\alpha }^3=0$  thus locus of P is  $y^3+(2-x)y{\alpha }^2-y{\alpha }^3=0$
$\Rightarrow y^2+(2-x){\alpha }^2-{\alpha }^3=0,\text{\hspace{0.17em}as\hspace{0.17em}}y\ne 0$  (P does not lie on the axis of the parabola)
 $\Rightarrow y^2={\alpha }^{2}x-2{\alpha }^2+{\alpha }^3$
If it is a part of the parabola  $y^2=4x,$ then