If a pair of lines x^{2}-2 p x y-y^{2}=0 and | vedclass

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(A) The equation of the pair of angle bisectors of the pair of lines $ax^{2}+2hxy+by^{2}=0$ is given by $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.For t

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$p q=-1$

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(A) The equation of the pair of angle bisectors of the pair of lines $ax^{2}+2hxy+by^{2}=0$ is given by $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.For the pair $x^{2}-2pxy-y^{2}=0$,we have $a=1, h=-p, b=-1$.The equation of the angle bisectors is $\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^{2}-y^{2}}{2} = \frac{xy}{-p}$.This gives $x^{2}-y^{2} = -\frac{2}{p}xy$,or $x^{2} + \frac{2}{p}xy - y^{2} = 0$.It is given that this pair is the same as $x^{2}-2qxy-y^{2}=0$.Comparing the coefficients of $xy$,we get $-2q = \frac{2}{p}$.Therefore,$pq = -1$.

If a pair of lines $x^{2}-2 p x y-y^{2}=0$ and $x^{2}-2 q x y-y^{2}=0$ are such that each pair bisects the angle between the other pair,then

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