# parabola locus problem

If $Q_1$ and $Q_2$ be the angle made by tangents to the axis of $y^2=4x$ from point $P$ and if $Q_1+Q_2=45^{\circ}$ then locus of $P$ is for options see here

• – lab bhattacharjee Mar 24 '18 at 17:51
• Please make your questions self-contained instead of making the people you’re asking for help go chasing links that can go stale. – amd Mar 24 '18 at 23:53
• @amd the link which I added is just additional information which is not at all necessary for solving the problem.... nevertheless I will take care next time – shreyans jain Apr 6 '18 at 3:21

Let $P(X,Y)$ be the pole.
$$Yy-2(x+X)=0 \tag{1}$$
Substitute $x=\dfrac{y^2}{4}$ into $(1)$,
\begin{align} y^2-2Yy+2X &= 0 \\ y_1+y_2 &= 2Y \\ y_1 y_2 &= 4X \\ m_1 &= \frac{2}{y_1} \tag{$2yy'=4$} \\ m_2 &= \frac{2}{y_2} \\ \frac{m_1+m_2}{1-m_1 m_2} &= \frac{2(y_1+y_2)}{y_1 y_2-4} \\ \tan 45^{\circ} &= \frac{4Y}{4X-4} \\ Y &= X-1 \end{align}