If the coordinate axes are the bisectors of t | vedclass

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

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$h = 0$

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(B) The equation of the pair of angle bisectors of the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.Since the coordinate axes are the bisectors,their equations are $x = 0$ and $y = 0$,which implies their combined equation is $xy = 0$.Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we see that the coefficient of $x^2$ and $y^2$ in the bisector equation must be zero.This implies $a - b = 0$ is not possible as per the given condition $a 
eq b$,or rather,for the bisectors to be the axes,the $x^2$ and $y^2$ terms must vanish in the bisector equation.Actually,the condition for the coordinate axes to be the bisectors is $a + b = 0$ and $h 
eq 0$ is not required,but rather $h$ must be such that the equation reduces to $xy = 0$.Given the standard form,if $a + b = 0$,the bisectors are $x^2 - y^2 = 0$,which are $y = \pm x$. If the axes are the bisectors,then $h$ must be $0$ and $a+b=0$ is not the condition; rather,the condition is $a+b=0$ is for perpendicular lines. For axes to be bisectors,$h=0$ is the condition.

If the coordinate axes are the bisectors of the angles between the pair of lines $ax^2 + 2hxy + by^2 = 0$,where $h^2 > ab$ and $a \neq b$,then

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