If the line y = mx bisects the angle between | vedclass

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

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$hx^2 + (a - b)x - h = 0$

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(A) The equation of the bisectors of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.This can be rewritten as $h(x^2 - y^2) - (a - b)xy = 0 \dots (1)$.Since the line $y = mx$ is a bisector,it must satisfy equation $(1)$.Substituting $y = mx$ into equation $(1)$,we get:$h(x^2 - (mx)^2) - (a - b)x(mx) = 0$$h(x^2 - m^2x^2) - (a - b)mx^2 = 0$Dividing by $x^2$ (assuming $x 
eq 0$):$h(1 - m^2) - (a - b)m = 0$$h - hm^2 - (a - b)m = 0$$hm^2 + (a - b)m - h = 0$.Thus,$m$ is a root of the quadratic equation $hx^2 + (a - b)x - h = 0$.

If the line $y = mx$ bisects the angle between the lines $ax^2 + 2hxy + by^2 = 0$,then $m$ is a root of the quadratic equation:

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