If the pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ be such that each pair bisects…
If the pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ be such that each pair bisects the angle between the other pair, then
$p q=1$
$p q=-1$
$p q=2$
$p q=-2$
Solution
Given that, the pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ bisects the angle between other pairs.
Since, we know that the pair of bisectors of the angle between the pair of straight lines
$a x^2+2 h x y+b y^2=0$ is $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
Hence, for $x^2-2 p x y-y^2=0$, the pair of bisectors will be
$\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$
$\Rightarrow \quad-p\left(x^2-y^2\right)=2 x y$
$\begin{aligned} & \Rightarrow \quad p x^2+2 x y-p y^2=0 \\ & \Rightarrow \quad x^2+\frac{2}{p} x y-y^2=0\end{aligned}$
But $x^2-2 q x y-y^2$ is the pair of bisectors.
So, $\frac{q}{p}=-2 q$
$\begin{aligned} & \Rightarrow-p q=1 \\ & \Rightarrow \quad p q=-1\end{aligned}$