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(C) Let the coordinates of $P$ be $(x, y)$. According to the given condition,the distance from $P(x, y)$ to $A(1, 1)$ equals the distance from $P(x, y

Vedclass · Accepted Answer

a parabola

Vedclass · Answer

(C) Let the coordinates of $P$ be $(x, y)$. According to the given condition,the distance from $P(x, y)$ to $A(1, 1)$ equals the distance from $P(x, y)$ to the line $x+y+2=0$.$\sqrt{(x-1)^2+(y-1)^2} = \frac{|x+y+2|}{\sqrt{1^2+1^2}}$Squaring both sides,we get:$(x-1)^2+(y-1)^2 = \frac{(x+y+2)^2}{2}$$2(x^2-2x+1+y^2-2y+1) = x^2+y^2+4+2xy+4x+4y$$2x^2+2y^2-4x-4y+4 = x^2+y^2+2xy+4x+4y+4$$x^2+y^2-2xy-8x-8y = 0$This equation is of the form $ax^2+2hxy+by^2+2gx+2fy+c=0$ where $h^2-ab = (-1)^2 - (1)(1) = 0$. Since $h^2=ab$ and the point $A$ does not lie on the line $x+y+2=0$,the locus represents 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 the locus of $P$ is

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