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Q)

If the pair 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 then,

$\begin{array}{1 1}(A)\;pq=-1 \\(B)\;p=q \\(C)\;p=-q\\(D)\;pq=1 \end{array}$

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A)
Equation of bisects of second pair of straight lines is
$qx^2+2xy-qy^2=0$------(1)
It must be identical to the first pair
$x^2-2pxy-y^2=0$------(2)
From (1) and (2)
$\large\frac{q}{1} =\frac{2}{-2p}=\frac{-q}{1}$
=> $pq=-1$
Hence A is the correct answer.
Equation of bisects of second pair of straight lines is
qx2+2xy−qy2=0------(1)

How did you get it ? Please prove
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