If the bisectors of the pair of lines x^2-2 m | vedclass

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(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of

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$mn+1=0$

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(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$. Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{x y}{-m}$. This simplifies to $\frac{x^2-y^2}{2} = \frac{x y}{-m}$,which implies $-m(x^2-y^2) = 2 x y$. Rearranging the terms,we get $m x^2 + 2 x y - m y^2 = 0$. Dividing by $m$ (assuming $m 
eq 0$),we get $x^2 + \frac{2}{m} x y - y^2 = 0$. Comparing this with the given bisector equation $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$. Therefore,$-n = \frac{1}{m}$,which gives $mn = -1$ or $mn+1=0$.

If the bisectors of the pair of lines $x^2-2 m x y-y^2=0$ are represented by $x^2-2 n x y-y^2=0$,then

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