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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial