Question

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

Answer 1

We have $x^2-2pxy-y^2=0-------\left(1\right)$

Compare $\left(1\right)$ with $a{x}^2+2hxy+b{y}^2=0$

we get $a=1,2h=-2p,b=-1$

$\therefore h=-p$

This equation of the bisectors of the angles between the lines given by $\left(1\right)$ is

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

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

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

or $P{x}^2+2xy-p{y}^2=0---------------\left(2\right)$

But it is given that,

$x^2-2qxy-y^2=0------------------\left(3\right)$

Is the equation of the bisectors of the angles between the line $\left(1\right)$

$\therefore \left(2\right)and\left(3\right)$ are the same.

$\therefore \frac{p}{1}=\frac{-p}{-1}$

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

$\Rightarrow pq=-1$