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4PM1/02 Jan 2020, Question 9: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 16 marks

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The question, in short

Given curve y = (qx − 2)/(x − p), with y-intercept A and normal y = x + 2 at A, determine p and q, sketch the curve, and find where the normal meets the curve again.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 7 marks

Show that p equals 1, then determine the numerical value of q using the normal condition at A.

  1. 1l meets the y-axis at (0, 2), so A = (0, 2)
  2. 2At x = 0: y = −2/(−p) = 2/p = 2, so p = 1
  3. 3dy/dx = (q(x − 1) − (qx − 2))/(x − 1)² = (2 − q)/(x − 1)²
  4. 4Normal gradient 1, so tangent gradient at A is −1
  5. 5At x = 0: 2 − q = −1, so q = 3

Answerp = 1 shown; q = 3

Part (b) · 5 marks

Draw the curve on the supplied axes, marking asymptotes and the coordinates where it cuts both axes.

  1. 1Vertical asymptote x = 1; horizontal asymptote y = 3
  2. 2x = 0 gives y = 2; y = 0 gives x = 2/3
  3. 3y = 3 + 1/(x − 1), so the branches lie top-right and bottom-left of (1, 3)
  4. 4Draw both branches approaching but not touching the asymptotes

AnswerTwo-branch rectangular hyperbola y = (3x − 2)/(x − 1) with asymptotes x = 1 and y = 3; crosses the y-axis at (0, 2) and the x-axis at (2/3, 0); branches lower-left and upper-right of the asymptotes' intersection

Part (c) · 4 marks

The normal line intersects the curve at a second point D; determine D's x coordinate.

  1. 1(3x − 2)/(x − 1) = x + 2
  2. 23x − 2 = (x + 2)(x − 1) = x² + x − 2
  3. 3x² − 2x = 0, so x(x − 2) = 0
  4. 4x = 0 is A, so D has x = 2

Answerx = 2

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