The tangents to the parabola $y^2=4 a x$ from an external point $P$ make angles $ heta_1$ and $ heta_2$ with the axis of the parabola. Such that $ an heta_1+ an heta_2=b$, where $b$ is constant. Then $P$ lies on

Question

The tangents to the parabola $y^2=4 a x$ from an external point $P$ make angles $	heta_1$ and $	heta_2$ with the axis of the parabola. Such that $	an 	heta_1+	an 	heta_2=b$, where $b$ is constant. Then $P$ lies on

Aicharya · Accepted Answer

Here, $P$ is intersecting point of tangents at $ \begin{gathered} A\left(a t_1^2, 2 a t_1\right) \text { and } B\left(a t_2^2, 2 a t_2\right) \\ P\left(a t_1 t_2 a\left(t_1 t_2\right)\right. \end{gathered} $

Slope of line $P A$ $ =\tan \theta_1=\frac{2 a t_1-a t_1-a t_2}{a t_1^2-a t_1 t_2}=\frac{a\left(t_1-t_2\right)}{a t_1\left(t_1-t_2\right)}=\frac{1}{t_1} $ Similarly, Slope of line $P B=\tan \theta_2=\frac{2 a t_2-a t_1-a t_2}{a t_2^2-a t_1 t_2}$ $ =\frac{a\left(t_2-t_1\right)}{a t_2\left(t_2-t_1\right)}=\frac{1}{t_2} $ According to the question, $ \begin{aligned} & \tan \theta_1+\tan \theta_2=b \\ & \therefore \quad \frac{1}{t_1}+\frac{1}{t_2}=b \\ & t_2+t_1=b t_1 t_2 \\ & \frac{y}{a}=\frac{b x}{a} \quad\left[\because x=a t_1 t_2, y=a\left(t_1+t_2\right)\right] \\ & y=b x \end{aligned} $ $\therefore \quad P$ lies on the line $y=b x$

The tangents to the parabola $y^2=4 a x$ from an external point $P$ make angles $\theta_1$ and $\theta_2$…

Solution