The combined equation of the bisectors of the angles between the lines joining the origin to the points of…
The combined equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+y^2+x y+x+3 y+1=0$ and the line $x+y+2=0$ is
$x^2+4 x y-y^2=0$
$x^2-4 x y-y^2=0$
$x^2-3 x y+y^2=0$
$x^2+2 x y-3 y^2=0$
Solution
Given equation: $x^2+y^2+x y+x+3 y+1=0$ ...(i)
and $x+y+2=0 \Rightarrow \frac{x+y}{-2}=1$ ...(ii)
Homogenizing (i) with (ii)
$\begin{aligned}
& x^2+x y+y^2+x(1)+3 y(1)+1(1)^2=0 \\
& \Rightarrow x^2+x y+y^2+x\left(\frac{x+y}{-2}\right)+3 y\left(\frac{x+y}{-2}\right)+\left(\frac{x+y}{-2}\right)^2=0 \\
& \Rightarrow 3 x^2-2 x y-y^2=0
\end{aligned}$ Comparing with $a x+2 h x y+b y^2=0$
$a=3,2 h=-2, b=-1$ Angle bisector is $h\left(x^2-y^2\right)=(a-b) x y$
$\begin{aligned}
& \Rightarrow-1\left(x^2-y^2\right)=(3+1) x y \\
& \Rightarrow x^2+4 x y-y^2=0 .
\end{aligned}$