A line passing through $P\left(2,3\right)$ and making an angle of $30°$ with the positive direction of $x$-axis meets $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA.PB$ is

Given,

Equation of the line through $P\left(2,3\right)$ is

So parametric equation of line is given by,

$\frac{x-2}{\cos \theta }=\frac{y-3}{\sin \theta }=r \text{where} \theta =30°$

Now if $PA=r_1, PB=r_2$ then $r_1,r_2$ are the roots of the equation

$x^2-2xy-y^2=0$
$\Rightarrow {\left(2+r\cos \theta \right)}^2-2\left(2+r\cos \theta \right)\left(3+r\sin \theta \right)-{\left(3+r\sin \theta \right)}^2=0$

$\Rightarrow r^2\left(\cos 2\theta -\sin 2\theta \right)-2r\left(\cos \theta +5\sin \theta \right)-17=0$

So $PA·PB=r_1{r}_2=\frac{-17}{\cos 2\theta -\sin 2\theta }=17\left(\sqrt{3}+1\right)$