0606/12 Feb/Mar 2025, Question 10: Tangents, Normals and Stationary Points (Further)
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 9 marks
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The question, in short
For y = (5x + 2)^(2/3) at x = 5, find where the normal meets x + y = 11 at Q, then reflect Q in the tangent at P to get R's coordinates.
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Solution · 9 marks
- 1At x = 5: y = 27^(2/3) = 9, so P = (5, 9)
- 2dy/dx = (2/3)(5x + 2)^(−1/3) × 5
- 3At x = 5: dy/dx = (10/3)(1/3) = 10/9
- 4Normal: y − 9 = −(9/10)(x − 5)
- 5With y = 11 − x: 2 − x = −(9/10)(x − 5), giving x = −25, y = 36, so Q = (−25, 36)
- 6Q lies on the normal, which is perpendicular to the tangent at P, so P is the midpoint of QR
- 7R = (2 × 5 − (−25), 2 × 9 − 36) = (35, −18)
Answer(35, −18)
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