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Which points on the graph of y=x^2 are closest to the point (0,1)

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Answer:

(±√(1/2), 1/2)

Explanation:

You want the points on the graph of y = x^2 that are closest to the point (0, 1).

Distance

The distance from the point (x, y) on the graph to the point (0, 1) is given by the distance formula. For the purpose here, minimizing the square of the distance will be equivalent to minimizing the distance.

d² = (x -0)² +(y -1)²

d² = x² +(x² -1)² = x² +x⁴ -2x² +1 = x⁴ -x² +1

This is a quadratic in x². It can be written in vertex form as ...

d² = (x² -1/2)² +3/4

This tells us two things:

  • distance is minimized when x² = 1/2
  • the minimum distance squared is 3/4

The points on the graph of y=x² closest to (0, 1) are ...

(x, y) = (±√(1/2), 1/2) . . . . . . . as shown in the attachment

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Additional comment

Effectively, we want the points of tangency for the largest circle centered at (0, 1) that is tangent to the curve y = x². The radius of that circle is √(3/4).

Which points on the graph of y=x^2 are closest to the point (0,1)-example-1
User Maher Aldous
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