If the pairs of straight lines x^2-2 q x y-y^ | vedclass

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(C) The equation of the 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^2-2px

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$p q+1=0$

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(C) The equation of the 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 angle bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^2-y^2}{2} = \frac{xy}{-p}$,or $x^2-y^2 = -\frac{2}{p}xy$,which is $x^2 + \frac{2}{p}xy - y^2 = 0$.It is given that this pair of bisectors is the same as the pair $x^2-2qxy-y^2=0$.Comparing the coefficients of $xy$,we get $\frac{2}{p} = -2q$.This implies $pq = -1$,or $pq+1=0$.Thus,option $C$ is correct.

If the pairs of straight lines $x^2-2 q x y-y^2=0$ and $x^2-2 p x y-y^2=0$ bisect the angles between each other,then which of the following is correct?

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