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4PM1/02 Jun 2023, Question 10: 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

For curve y = (6 − 3x)/(x − 4), x ≠ 4, find intercepts P and Q, asymptotes, sketch C, then find the normal L at x = 2 and where L meets C again at R.

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

Part (a) · 2 marks

Find the coordinates of P where the curve crosses the x-axis, and of Q where it crosses the y-axis.

  1. 1y = 0: 6 − 3x = 0, x = 2, so P(2, 0)
  2. 2x = 0: y = 6/(−4) = −3/2, so Q(0, −3/2)

Answer(i) P(2, 0) (ii) Q(0, −3/2)

Part (b) · 2 marks

State the equations of the two asymptotes: the one parallel to the y-axis and the one parallel to the x-axis.

  1. 1Vertical asymptote where x − 4 = 0: x = 4
  2. 2As x → ±∞, y → −3x/x = −3

Answer(i) x = 4 (ii) y = −3

Part (c) · 3 marks

Sketch C, clearly showing both asymptotes and the coordinates of P and Q.

  1. 1Draw dashed asymptotes x = 4 and y = −3
  2. 2Left branch rises from near y = −3 through (0, −3/2) and (2, 0) up towards x = 4
  3. 3Right branch comes up from below near x = 4 towards y = −3

AnswerTwo branches with asymptotes x = 4 and y = −3: one branch top-left (above y = −3, left of x = 4) passing through Q(0, −3/2) and P(2, 0); the other bottom-right (below y = −3, right of x = 4)

Part (d) · 6 marks

The line L is normal to C at the point where x = 2. Find an equation of L.

  1. 1dy/dx = [−3(x − 4) − (6 − 3x)]/(x − 4)² = 6/(x − 4)²
  2. 2At x = 2: dy/dx = 6/4 = 3/2, and y = 0
  3. 3Gradient of normal = −2/3
  4. 4y − 0 = −(2/3)(x − 2)
  5. 53y = −2x + 4

Answer3y = −2x + 4

Part (e) · 3 marks

Given L meets C again at R, find the x coordinate of R.

  1. 1(6 − 3x)/(x − 4) = (−2x + 4)/3
  2. 23(6 − 3x) = (x − 4)(−2x + 4), so 2x² − 21x + 34 = 0
  3. 3(2x − 17)(x − 2) = 0
  4. 4x = 2 is the given point, so at R, x = 17/2

Answerx = 17/2

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