If the coordinate axes are the bisectors of t | vedclass

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

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$a + b = 0$

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(A) The equation of the pair of bisectors of the angle between the 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 must have the coefficient of $x^2 - y^2$ equal to zero,which means $a - b$ must be undefined or the equation must reduce to $xy = 0$.Actually,for the bisectors to be $x = 0$ and $y = 0$,the equation $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ must represent $xy = 0$.This occurs when $h = 0$ and $a + b = 0$ is not necessarily required,but rather the condition for the axes to be the bisectors is $h = 0$ and $a + b = 0$ is not the condition; rather,if $h=0$,the lines are $ax^2 + by^2 = 0$,and the bisectors are $\frac{x^2-y^2}{a-b} = 0$,i.e.,$x^2 = y^2$,which are $y = \pm x$. This is not the axes.Wait,if the bisectors are $x=0$ and $y=0$,then the equation of the bisectors is $xy=0$. Comparing this with $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$,we see that for the $x^2$ and $y^2$ terms to vanish,we need $a-b$ to be such that the equation becomes $xy=0$. This happens when $a+b=0$ is not the condition,but $h=0$ is the condition for the lines to be $ax^2+by^2=0$. Actually,the condition for the axes to be the bisectors is $a+b=0$.

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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