Q90 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If the equation of pair of lines $y=m_1 x$ and $y=m_2 x$ can be written as $\left(y-m_1 x\right)$ $\left(y-m_2 x\right)=0$. Then the equation of pair of the angle bisector of the lines $5 x^2+2 x y+3 y^2=0$ is
- A$x^2-4 x y-y^2=0$
- Correct answerB$x^2-2 x y-y^2=0$
- C$x^2-8 x y-y^2=0$
- D$x^2-6 x y-y^2=0$
Solution
If $L \equiv a x^2+b y^2+2 h x y=0$
Then equation of pair of angle bisector is $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
$\therefore$ Equation of pair of angle bisector of $5 x^2+2 x y+3 y^2=0$ is
$\begin{array}{rlrl} <br/>& & \frac{x^2-y^2}{5-3} & =\frac{x y}{1} \\<br/>\Rightarrow & & x^2-y^2 & =2 x y \\<br/>\Rightarrow & x^2-2 x y-y^2 & =0<br/>\end{array}$
Then equation of pair of angle bisector is $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
$\therefore$ Equation of pair of angle bisector of $5 x^2+2 x y+3 y^2=0$ is
$\begin{array}{rlrl} <br/>& & \frac{x^2-y^2}{5-3} & =\frac{x y}{1} \\<br/>\Rightarrow & & x^2-y^2 & =2 x y \\<br/>\Rightarrow & x^2-2 x y-y^2 & =0<br/>\end{array}$
Hence, the correct answer is option: 'B'.