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

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(B) The equation of the pair of bisectors of the angle between 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}=\fr

Vedclass · Accepted Answer

$p q=-1$

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(B) The equation of the pair of bisectors of the angle between 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 $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.The equation of the bisectors is $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2xy$,or $p x^2+2 x y-p y^2=0$.Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.We are given that this pair is $x^2-2 q x y-y^2=0$.Comparing the coefficients of $xy$,we have $-2 q = \frac{2}{p}$.Therefore,$-p q = 1$,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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