If the coordinate axes are the bisectors of t | vedclass

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

Vedclass · Accepted Answer

$h 
eq 0, a + b = 0$

Vedclass · Answer

(C) The equation of the pair of angle bisectors of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.Since the coordinate axes are the bisectors,their equation is $xy = 0$.Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we observe that the $x^2$ and $y^2$ terms must vanish.This implies $a - b = 0$,which contradicts the given condition $a 
eq b$.However,if we look at the structure $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$,for this to represent $xy = 0$,the coefficient of $xy$ in the bisector equation must be non-zero,and the $x^2 - y^2$ term must be zero.Actually,the condition for the coordinate axes to be the bisectors is $a + b = 0$ and $h 
eq 0$.

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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