If a point $P$ moves such that its distances from the point $A(1,1)$ and the line $x+y+2=0$ are equal, then the locus of $P$ is

Question

Aicharya · Accepted Answer

Let the co-ordinates of $P$ are $(x, y)$ According to given condition $\begin{aligned} & (x-1)^2+(y-1)^2=\frac{(x+y+2)^2}{2} \ & \Rightarrow \quad 2\left(x^2+1-2 xight)+2\left(y^2+1-2 yight) \ & \quad=x^2+y^2+4+2 x y+4 y+4 x \end{aligned}$ On comparing with $\begin{aligned} & a x^2+b y^2+2 h x y+2 g x+2 f y+c=0 	ext {, we get } \ & a=1, b=1, h=-1, g=-4, f=-4, c=0 \ & 	herefore a b c+2 f g h-a f^2-b g^2-c h^2 \ & =1.1 .0+2(-4)(-4)(-1)-1(-4)^2-1(-4)^2-0 \ & =0-32-16-16=-64 
eq 0 \ & 	ext { and } h^2-a b=1-1=0 \ & 	ext { Since } \Delta 
eq 0 	ext { and } h^2=a b 	ext {, } \ & \end{aligned}$ therefore locus of $P$ is a parabola.

If a point $P$ moves such that its distances from the point $A(1,1)$ and the line $x+y+2=0$ are equal, then…

Solution