Line passing through the point $P (2, 3)$ meets the lines represented by$x^2-2xy-y^2=0$ at the points A and B such that $PA.PB=17$, the equation of the line is

Solution:

Let the equation of the line through$P(2, 3)$ making an angle $\theta$ with the positive direction of x-axis be

$\frac{x-2}{\cos \theta }=\frac{y-3}{\sin \theta }$.

Then the coordinates of any point on this line at a distance 

r from p are $(2+r\cos \theta ,3+r\sin \theta )$. If $PA=r_1$ and $PB=r_2$, then $r_1,r_2$ are the roots of the equation.

$\begin{array}{l}(2+r\cos \theta {)}^2-2(2+r\cos \theta \left)\right(3+r\sin \theta )-(3+r\sin \theta {)}^2=0\\ \Rightarrow r^2(\cos 2\theta -\sin 2\theta )-2r(\cos \theta +5\sin \theta )-17=0\end{array}$

$\Rightarrow 17=\mathbf{PA}\cdot \mathbf{PB}=r_1{r}_2=\frac{17}{\cos 2\theta -\sin 2\theta }$

$\Rightarrow \cos 2\theta -\sin 2\theta =1$ which is satisfied by $\theta =0$and thus
the equation of the line is$y = 3$.