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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

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  • $\begingroup$ See also: math.stackexchange.com/questions/2600622/… $\endgroup$ – lab bhattacharjee Mar 24 '18 at 17:51
  • $\begingroup$ Please make your questions self-contained instead of making the people you’re asking for help go chasing links that can go stale. $\endgroup$ – amd Mar 24 '18 at 23:53
  • $\begingroup$ @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 $\endgroup$ – shreyans jain Apr 6 '18 at 3:21
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Let $P(X,Y)$ be the pole.

The equation of polar or equivalently the chord:

$$Yy-2(x+X)=0 \tag{1}$$

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}

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