Question

If $y=mx$ be one of the bisectors of the angle between the lines $a{x}^2-2hxy+b{y}^2=0$, then

• A. $h\left(1+m^2\right)+m\left(a-b\right)=0$
• B. $h\left(1-m^2\right)+m\left(a+b\right)=0$
• C. $h\left(1-m^2\right)+m\left(a-b\right)=0$
• D. $h\left(1+m^2\right)+m\left(a+b\right)=0$

Answer 1

The correct option is C

$h\left(1-m^2\right)+m\left(a-b\right)=0$

Explanation for the correct option.

Find the correct relation.

It is known that for lines given by the equation $a{x}^2+2hxy+b{y}^2=0$, the equations of angle bisector are given as: $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$.

It is given that the equation of the lines is $a{x}^2-2hxy+b{y}^2=0$, so $h$ is replaced by $-h$ and thus the equations of the angle bisector are $\frac{x^2-y^2}{a-b}=\frac{xy}{-h}$.

Upon cross multiplying we get

$\begin{array}{rcl}-h\left(x^2-y^2\right)& =& xy\left(a-b\right)\\ & \Rightarrow & h\left(x^2-y^2\right)+xy\left(a-b\right)=0\end{array}$

Now, it is given that $y=mx$ is one of the angle bisector so substitute $mx$ for $y$.

$$\begin{gathered} h\left(x^2-{\left(mx\right)}^2\right)+x\times mx\left(a-b\right)=0 \\ \Rightarrow x^2\left[h\left(1-m^2\right)+m\left(a-b\right)\right]=0 \\ \Rightarrow h\left(1-m^2\right)+m\left(a-b\right)=0 \end{gathered}$$

So the relation is $h\left(1-m^2\right)+m\left(a-b\right)=0$.

Hence, the correct option is C.