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NDA 2025 Maths · Mathematics · Application of Derivatives · hard

The least value of the square of the distance of a point on the curve y=x^2 from the point (0,2) is

Single correct option

The least value of the square of the distance of a point on the curve y=x2y=x^2 from the point (0,2)(0,2) is
  1. A. 22
  2. B. 32\dfrac32
  3. C. 44
  4. D. 74\dfrac74
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Answer: (D) 74\dfrac74

Let D=x2+(x2−2)2=x4−3x2+4D=x^2+(x^2-2)^2=x^4-3x^2+4. Then D′=4x3−6x=2x(2x2−3)D'=4x^3-6x=2x(2x^2-3), giving x2=32x^2=\frac32 (the point x=0x=0 gives the maximum 44 locally). So D=94−92+4=74D=\frac94-\frac92+4=\frac74. The value 44 is the distance squared at x=0x=0, which is a local maximum of DD, not the minimum.
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