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

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Question

(B) The equation of the pair of angle bisectors for the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$

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$p q=-1$

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(B) The equation of the pair of angle bisectors for the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.For the pair $x^2-2 p x y-y^2=0$,the bisectors are given by $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2 x y$,or $p x^2+2 x y-p y^2=0$.Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.Since this is the pair of bisectors,it must be identical to the given pair $x^2-2 q x y-y^2=0$.Comparing the coefficients of $x y$,we have $-2 q = \frac{2}{p}$.Thus,$-2 p q = 2$,which gives $p q = -1$.

If the pairs of straight 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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