A point P(x,y) is such that its distance from the fixed point `(alpha,0)` is equal to its distance from y-axis. Prove that the equation of the locus is given by, `y^2 = alpha (2x-alpha)`.

Explanation

It is given that `bar(AP) = bar(BP)`
or, `sqrt((x-alpha)^2 + y^2) = x`
or, `(x-alpha)^2 + y^2= x^2`
or, `y^2 = x^2-x^2-alpha^2 + 2alphax = alpha(2x-alpha)`
`therefore y^2 = alpha(2x-alpha)`
which is the locus of the point P(x,y).