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4PM1/02R Jun 2024, Question 7: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 14 marks

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The question, in short

For the curve y = (x² − 1)/(4x + 5), x ≠ −5/4, identify its vertical asymptote, then work with the normal at x = −1 and its second crossing with the curve.

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

Part (a) · 1 mark

State the equation of the vertical asymptote of the curve.

  1. 14x + 5 = 0
  2. 2x = −5/4

Answerx = −5/4

Part (b) · 7 marks

Determine an equation for the normal line l to the curve at the point where x = −1.

  1. 1dy/dx = [2x(4x + 5) − 4(x² − 1)]/(4x + 5)²
  2. 2At x = −1: y = 0
  3. 3dy/dx = [−2(1) − 0]/1 = −2
  4. 4Gradient of normal = 1/2
  5. 5y − 0 = ½(x + 1)

Answery = ½(x + 1) (2y = x + 1)

Part (c) · 6 marks

Calculate the coordinates of point D, where line l intersects the curve again.

  1. 1(x + 1)/2 = (x² − 1)/(4x + 5)
  2. 2(x + 1)(4x + 5) = 2(x² − 1)
  3. 32x² + 9x + 7 = 0, (2x + 7)(x + 1) = 0
  4. 4x = −7/2 (x = −1 is the original point)
  5. 5y = (−7/2 + 1)/2 = −5/4

AnswerD(−7/2, −5/4)

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