If the pair of straight lines x^2 - 2mxy - y^ | vedclass

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(B) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{

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(B) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.For the first pair $x^2 - 2mxy - y^2 = 0$,we have $a = 1, h = -m, b = -1$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-m}$ $\Rightarrow \frac{x^2 - y^2}{2} = \frac{xy}{-m}$ $\Rightarrow -m(x^2 - y^2) = 2xy$ $\Rightarrow mx^2 + 2xy - my^2 = 0$.For the second pair $x^2 - 2nxy - y^2 = 0$,we have $a = 1, h = -n, b = -1$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-n} \Rightarrow nx^2 + 2xy - ny^2 = 0$.Since each pair bisects the angle between the other,the bisectors of the first pair must be the second pair,and vice versa. Comparing $mx^2 + 2xy - my^2 = 0$ with $x^2 - 2nxy - y^2 = 0$,we get $\frac{m}{1} = \frac{2}{-2n} = \frac{-m}{-1}$.From $\frac{m}{1} = \frac{2}{-2n}$,we get $m = -\frac{1}{n}$,which implies $mn = -1$.

If the pair of straight lines $x^2 - 2mxy - y^2 = 0$ and $x^2 - 2nxy - y^2 = 0$ are such that each pair bisects the angle between the other pair,then $mn =$

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