Question

find the equation of locus of a point equidistant from A(2,0) and the y axis.

Answer 1

Answer:  The required locus of the point is $4x-2by+b^2=4.$

Step-by-step explanation:  We are given to find the equation of the locus of a point that is equidistant from A(2, 0) and the Y-axis.

Let (x, y) represent the point that is equidistant from A(2, 0) and the Y-axis.

Any point on Y-axis can be written as (0, b), because the x co-ordinate is zero on the X-axis.

Distance formula :  The distance between the points (p, q) and (r, s) is given by

$D=\sqrt{(r-p)^2+(s-q)^2}.$

According to the given information, we have

$\sqrt{(x-2)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-b)^2}\\\\\Rightarrow \sqrt{(x-2)^2+y^2}=\sqrt{x^2+(y-b)^2}\\\\\Rightarrow (x-2)^2+y^2=x^2+(y-b)^2~~~~~~~~~~~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow x^2-4x+4+y^2=x^2+y^2-2by+b^2\\\\\Rightarrow -4x+4+2by-b^2=0\\\\\Rightarrow 4x-2by+b^2=4.$

Thus, the required locus of the point is $4x-2by+b^2=4.$

Answer 2

here is youre answer it it correct