Question

If a pair of lines $x^2-2pxy-y^2=0$ and $x^2-2pxy-y^2=0$ is such that each pair bisects the angle between the other pair, prove that $pq=-1$.

Answer 1

Equations of the bisectors of $x^2-2pxy-y^2=0$

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

or $x^2-y^2=\frac{-2xy}{p}$ ..$\left(1\right)$

But the bisectors of first pair of lines are given by the $2$nd paid, i.e., by
$x^2-y^2=2qxy$ ..$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, we get

$-1/p=q$ or $pq=1$

or $pq+1=0$.