Question

Find the equation of Locus of a point equidistant from A (2,0) and Y-axis. ​

Answer 1

the equation of locus of a point equidistany from A(2,0) and Y-axis is y^2=4(x-1).

Answer 2

Step-by-step explanation:

Given :-

The point A(2,0)

To find :-

Equation of the locus of the point A which is equidistant from Y-axis

Solution :-

Given point A = (2,0)

We know that

Equation of Y-axis is X = 0

Let the point which is equidistant from the point A and Y-axis be P(x,y)

The distance from Y-axis to the point P(x,y) = x

A__________P_________Y-axis

The distance between A and P = The distance between P and Y-axis

We know that

The distance between two points (x₁, y₁) and (x₂, y₂) is √[(x₂-x₁)²+(y₂-y₁)²] units

Let (x₁, y₁) = (2,0) => x₁ = 2 and y₁ = 0

Let (x₂, y₂) = (x,y) => x₂ = x and y₂ = y

The distance between A and P

=> AP = √[(x-2)²+(y-0)²]

=> AP = √[(x-2)²+y²]

=> AP = √(x²-4x+4+y²) units -----------(1)

and

The distance from Y-axis to the point P(x,y)

= x ----------(2)

According to the given problem

The distance between A and P = The distance between P and Y-axis

=> √(x²-4x+4+y²) = x

On squaring both sides then

=> [√(x²-4x+4+y²)]² = x²

=> x²-4x+4+y² = x²

=> x²-4x+4+y²-x² = 0

=> -4x+4+y² = 0

=> y²-4x+4 = 0

Therefore,

Required equation is y²-4x+4 = 0

Answer :-

The equation of the locus of the point which is equidistant from A (2,0) and Y-axis is y²-4x+4 = 0

Used formulae:-

→ The distance between two points

(x₁, y₁) and (x₂, y₂) is √[(x₂-x₁)²+(y₂-y₁)²] units

→ The equation of Y-axis is x = 0