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

• A

  a straight line
• B

  a pair of straight lines
• Correct answer
  C

  a parabola
• D

  an ellipse

Solution

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

Hence, the correct answer is option: 'C'.