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If a pair of lines $x^{2}-2 p x y-y^{2}=0$ and $x^{2}-2 q x y-y^{2}=0$ is such that each pair bisects the angle between the other pair, then
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Verified Answer
The correct answer is:
$p q=-1$
Since, $x^{2}+\frac{2 x y}{p}-y^{2}=0$, is the angle bisectors of $x^{2}-2 p x y-y^{2}=0$
But given that angle bisectors are
$\therefore$
$$
\begin{array}{c}
x^{2}-2 q x y-y^{2}=0 \\
-2 q=2 / p
\end{array}
$$
$$
\Rightarrow \quad p q=-1
$$
But given that angle bisectors are
$\therefore$
$$
\begin{array}{c}
x^{2}-2 q x y-y^{2}=0 \\
-2 q=2 / p
\end{array}
$$
$$
\Rightarrow \quad p q=-1
$$
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