What is the minimum distance between the point (3, 0) and the parabola y^2 = 4x + 2?

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The minimum distance between the parabola y^2 = 4x + 2 and the point (3, 0) has to be found.

Let the point on the parabola that has the smallest distance from (3, 0) be (X, Y)

The square of the distance between the points is D = (X - 3)^2 + (Y - 0)^2. We have to determine values of X and Y such that D is minimized.

D = (X - 3)^2 + (Y - 0)^2

=> (X - 3)^2 + Y^2

Also, y^2 = 4x + 2

=> (X - 3)^2 + 4X + 2

D' = 2(X - 3) + 4

The value of D' is minimum at 2(X - 3) + 4 = 0

=> X - 3 = -4/2

=> X = -2 + 3

=> X = 1

D = (1 - 3)^2 + 4 + 2 = 10

The minimum distance between the point (3, 0) and the parabola y^2 = 4x + 2 is sqrt 10

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