Let A, P, B are collinear points on lines $y = 0, y = 2x, y = 3x$ respectively. If $PA.PB$ is minimum (for a fixed P), then which of the following is/are TRUE?

$$\begin{gathered} PA.PB=\frac{PM.PN}{\cos \varphi \cos (\theta -\varphi )} \\ \Delta =\frac{2PM.PN}{\cos \theta +\cos (2\varphi -\theta )} \end{gathered}$$

For PA.PB to be minimum  $2\varphi -\theta =0\Rightarrow \theta =2\varphi$
$\therefore OA=OB\Rightarrow \Delta OAB$  is isosceles
Slope of  $PA=\tan ({90}^0+\varphi )=-\cot \varphi =-\cot \frac{\theta }{2}$
$$\begin{gathered} \tan \theta =3\Rightarrow \frac{2t}{1-t^2}=3\Rightarrow 3-3t^2=2t \\ 3t^2+2t-3=0 \\ t=\frac{-2\pm \sqrt{4+36}}{6}=\frac{-1\pm \sqrt{10}}{3} \end{gathered}$$