Question

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

• A. $pq=-1$
• B. $pq=1$
• C. $\frac{1}{p}+\frac{1}{q}=0$
• D. $\frac{1}{p}-\frac{1}{q}=0$

Answer 1

The correct option is A $pq=-1$

Since, $x^2+\frac{2xy}{p}-y^2=0$ is the angle bisectors of $x^2-2pxy-y^2=0$
But given that angle bisectors are
$x^2-2qxy-y^2=0$
$\therefore -2q=\frac{2}{p}$
$\Rightarrow pq=-1$