If the lines represented by x^2-2hxy-y^2=0 ar | vedclass

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(C) The angle bisectors of two lines remain invariant under rotation of the lines by the same angle in opposite directions.Therefore,the bisectors of

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$hx^2-hy^2+2xy=0$

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(C) The angle bisectors of two lines remain invariant under rotation of the lines by the same angle in opposite directions.Therefore,the bisectors of the new lines are the same as the bisectors of the original lines $x^2-2hxy-y^2=0$.The equation of the angle bisectors for a pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.Here,$a=1$,$b=-1$,and the coefficient of $xy$ is $-2h$,so the coefficient of $xy$ in the formula is $2h' = -2h$,which means $h' = -h$.Substituting these values into the formula:$\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-h}$$\frac{x^2-y^2}{2} = \frac{xy}{-h}$$-h(x^2-y^2) = 2xy$$-hx^2+hy^2 = 2xy$$hx^2-hy^2+2xy = 0$

If the lines represented by $x^2-2hxy-y^2=0$ are rotated about $(0,0)$ through an angle $\alpha$,one in the clockwise direction and the other in the counter-clockwise direction,then the combined equation of the bisectors of the angle between the lines thus obtained is

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