If x^{2}-2 p x y-y^{2}=0 and x^{2}-2 q x y-y^ | vedclass

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Question

(C) The equation of the bisectors of the angles between the lines $x^{2}-2 p x y-y^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}

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$p q+1=0$

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(C) The equation of the bisectors of the angles between the lines $x^{2}-2 p x y-y^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.Here,$a=1, b=-1, h=-p$.So,$\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$.$\Rightarrow \frac{x^{2}-y^{2}}{2} = \frac{-xy}{p}$.$\Rightarrow px^{2} + 2xy - py^{2} = 0$.Since this equation represents the same pair of lines as $x^{2}-2qxy-y^{2}=0$,we compare the coefficients:$\frac{p}{1} = \frac{2}{-2q} = \frac{-p}{-1}$.From $\frac{p}{1} = \frac{2}{-2q}$,we get $p = -\frac{1}{q}$,which implies $pq = -1$ or $pq+1=0$.

If $x^{2}-2 p x y-y^{2}=0$ and $x^{2}-2 q x y-y^{2}=0$ bisect angles between each other,then

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