If the line y=mx is one of the bisectors of x | vedclass

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(A) The given equation is $x^2+4xy-y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we get $a=1, h=2, b=-1$. The equation of the angle bisectors is give

Vedclass · Accepted Answer

$-1+\sqrt{5}$

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(A) The given equation is $x^2+4xy-y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we get $a=1, h=2, b=-1$. The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$. Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{2}$. $\frac{x^2-y^2}{2} = \frac{xy}{2} \implies x^2-xy-y^2=0$. Since $y=mx$ is a bisector,substitute $y=mx$ into the equation: $x^2-x(mx)-(mx)^2=0$. $x^2(1-m-m^2)=0$. For non-trivial lines,$m^2+m-1=0$. Using the quadratic formula,$m = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Therefore,$2m = -1 \pm \sqrt{5}$.

If the line $y=mx$ is one of the bisectors of $x^2+4xy-y^2=0$,then the value of $2m$ is:

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