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

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

The explanation for the correct option:

Pair of straight line and angle-bisector:

It is known that the bisectors of the angle between the pair of straight lines $a{x}^2+2hxy+b{y}^2=0$ can be given by, $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$.

The given equation of the pair of straight lines, $a{x}^2-2hxy+b{y}^2=0$.

Thus, the equation of the bisectors of the angle between the given pair of straight lines is $\frac{x^2-y^2}{a-b}=\frac{xy}{-h}\_\left(1\right)$.

Now, it is given that one of the bisectors of the angle is $y=mx\_\left(2\right)$.

Thus, equation $\left(2\right)$ satisfies equation $\left(1\right)$.

So, $\frac{x^2-{\left(mx\right)}^2}{a-b}=\frac{x\left(mx\right)}{-h}$

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

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

Hence, (C) is the correct option.