If pairs of straight lines x^2-2 p x y-y^2=0 | vedclass

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(D) The equation of the angle bisectors for the pair of 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 first pa

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$p q=-1$

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(D) The equation of the angle bisectors for the pair of 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 first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.Given that this pair of bisectors is the same as the second pair $x^2-2 q x y-y^2=0$,we compare the coefficients.Comparing $x^2-y^2+\frac{2}{p} x y=0$ with $x^2-2 q x y-y^2=0$,we identify that the coefficient of $x y$ must match:$\frac{2}{p} = -2 q$.Therefore,$p q = -1$.

If 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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