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TMUA 2018 · Paper 1 · Question 16 of 20

TMUA 2018 Paper 1 Question 16

Coordinate geometry — Minimising a distance · quadratic in b2. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 1Coordinate geometryMinimising a distance · quadratic in b26 options
The curve C has equation y = x2 + bx + 2, where b ≥ 0.

Find the value of b that minimises the distance between the origin and the stationary point of the curve C.

  1. Ab = 0
  2. Bb = 1
  3. Cb = 2
  4. Db = √62
  5. Eb = √2
  6. Fb = √6
Show the answer and worked solution
answer · F
  1. Ab = 0
  2. Bb = 1
  3. Cb = 2
  4. Db = √62
  5. Eb = √2
  6. Fb = √6
Completing the square, the stationary point is at x = −b2 with y = 2 − b24. Minimise the squared distance rather than the distance itself: D2 = b24 + (2 − b24)2. Substituting u = b24 ≥ 0 turns this into u + (2−u)2 = u2 − 3u + 4, a quadratic in u minimised at u = 32. So b24 = 32, giving b2 = 6 and, since b ≥ 0, b = √6.