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
2026-03-25 11:08:03.1774436883
parabola locus problem
246 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
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}