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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
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Verified Answer
The correct answer is:
a parabola
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.
$\begin{aligned}
& (x-1)^2+(y-1)^2=\frac{(x+y+2)^2}{2} \\
& \Rightarrow \quad 2\left(x^2+1-2 x\right)+2\left(y^2+1-2 y\right) \\
& \quad=x^2+y^2+4+2 x y+4 y+4 x
\end{aligned}$
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.
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