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TMUA 2018 · Paper 1 · Question 11 of 20

TMUA 2018 Paper 1 Question 11

Differentiation and integration — Normals to a curve. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 1Differentiation and integrationNormals to a curve6 options
The line y = mx + 5, where m > 0, is normal to the curve y = 10 − x2 at the point (p,  q).

What is the value of p?

  1. A√26
  2. B−√26
  3. C3√22
  4. D−3√22
  5. E√5
  6. F−√5
Show the answer and worked solution
answer · C
  1. A√26
  2. B−√26
  3. C3√22
  4. D−3√22
  5. E√5
  6. F−√5
Since dydx = −2x, the tangent at (p,  q) has gradient −2p, so the normal has gradient m = 12p. The point lies on the line, so q = mp + 5 = 12p⋅ p + 5 = 12 + 5 = 112, and this holds whatever p is. It also lies on the curve, so 10 − p2 = 112, giving p2 = 92 and p = ±3√2. The condition m > 0 forces p > 0, so p = 3√22.