Question

Normals are drawn from the point $P$ with slopes $m_1,m_2,m_3$ to the parabola $y^2=4x.$ If the locus of $P$ when $m_1,m_2=\alpha$ is part of the parabola itself then the value of $\alpha$ is.

Answer 1

We know, equation of normal to $y^2=4ax$ is
$y=mx-2am-{am}^3$

Thus, equation of normal to $y^2=4x$ is,
$y=mx-2m-m^3$

Let it passes through $(h,k)$
$\Rightarrow k=mh-2m-m^3$

or $m^3+m(2-h)+k=0$ ...(i)

Here, $m_1+m_2+m_3=0,$

$m_1{m}_2+m_2{m}_3+m_3{m}_1=2-h$

$m_1{m}_2{m}_3=-k$ where $m_1{m}_2=\alpha$

$\Rightarrow m_3=-\frac{k}{\alpha }$ it must satisfy Eq. (i)

$\Rightarrow \frac{k^3}{{\alpha }^3}-\frac{k}{\alpha }(2-h)+k=0$

$\Rightarrow k^2={\alpha }^{2}h-{2\alpha }^2+{\alpha }^2$

$\Rightarrow y^2={\alpha }^{2}x-{2\alpha }^2+{\alpha }^3$

On comparing with $y^2=4x\Rightarrow {\alpha }^2=4$ and $-2{\alpha }^2+{\alpha }^3=0$

$\Rightarrow \alpha =2$