Question

Find the locus of the point which is equidistant
from A(2,0) and the y axis​

Answer 1

Step-by-step explanation:

Answer: The required locus of the point is 4x-2by+b^2=4.4x−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}.D=

(r−p)

2

+(s−q)

2

.

According to the given information, we have

Thus, the required locus of the point is 4x-2by+b^2=4.4x−2by+b

2

=4.