If the bisectors of the lines x^2 - 2pxy - y^ | vedclass

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Question

(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$

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$pq + 1 = 0$

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(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.For the given equation $x^2 - 2pxy - y^2 = 0$,we have $a = 1, h = -p, b = -1$.Substituting these values,the equation of the bisectors is $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-p}$.$\frac{x^2 - y^2}{2} = \frac{xy}{-p}$ $\Rightarrow -p(x^2 - y^2) = 2xy$ $\Rightarrow -px^2 + py^2 = 2xy$ $\Rightarrow px^2 + 2xy - py^2 = 0$.Comparing this with the given bisector equation $x^2 - 2qxy - y^2 = 0$,we have the ratio of 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 the bisectors of the lines $x^2 - 2pxy - y^2 = 0$ are $x^2 - 2qxy - y^2 = 0$,then

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