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

Dr. Abubacker · Accepted Answer

(c)
Let the coordinates of $P$ are $(x,\ y)$ according to given condition
$(x - 1)^{2} + (y - 1)^{2} = \frac{(x + y + 2)^{2}}{2}$
$\Rightarrow \ x^{2} + y^{2} - 2xy - 8x - 8y = 0$
Here, $a = 1,b = 1,h = - 1,\ g = - 4,\ f = - 4,\ c = 0$
Now , $abc + 2fgh - af^{2} - bg^{2} - ch^{2}$
$= 1.1.0 + 2( - 4)( - 4)( - 1) - 1( - 4)^{2} - 1( - 4)^{2} - 0$
$= - 64 
eq 0$
and $h^{2} - ab = 1 - 1 = 0$
Since, $\mathrm{\Delta} 
eq 0$ and $h^{2} = ab$
Hence, locus of $P$ is a parabola

Q. 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

Explanation

Reviewed by Dr. Abubacker

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