Question

If pairs of straight lines $x^2-2pxy-y^2=0$ and $x^2-2qxy-y^2=0$ be such that each pair bisects the angles between the other pair, prove that pq = -1.

Answer 1

Equation of angle bisector for $a{x}^2+2hxy+b{y}^2=0$ is

$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

For $x^2-2pxy-y^2=0$ Equation of angle bisector will be

$\Rightarrow \frac{x^2-y^2}{1-(-1)}=\frac{xy}{-p}$

$\Rightarrow x^2-y^2+\frac{2xy}{p}=0$

Given equation of angle bisector is $x^2-2qxy+y^2=0$

Comparing both the equations

$\Rightarrow \frac{2}{p}=-2q\Rightarrow pq=-1$

Hence proved.