Question

A line passes through the point $P(2,3)$ and makes an angle of ${30}^{\circ }$ with the positive direction of $x$axis. It meets the lines represented by $x^2-2xy-y^2=0$ at the points $A$ and $B$ such that $PA\cdot PB=a$, the value of $7(a-17{)}^2$ is equal to

Answer 1

Equation of the line through $P(2,3)$ is $\frac{x-2}{\cos \theta }=\frac{y-3}{\sin \theta }(\theta ={30}^{\circ })$
If $PA=r_1,PB=r_2$ then $r_1,r_2$ are the roots of the equation .
${(2+r\cos \theta )}^2-2(2+r\cos \theta )(3+r\sin \theta )-{(3+r\sin \theta )}^2=0$
$\Rightarrow r^2(\cos 2\theta -\sin 2\theta )-2r(\cos \theta +5\sin \theta )-17=0$
So $a=PA\cdot PB=r_1{r}_2=\frac{-17}{\cos 2\theta -\sin 2\theta }=17(\sqrt{3}+1)$
$7(a-17{)}^2=7\times 289\times 3=6069$