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1.6.3 Parametric equations of curves (A-level only)

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For a fixed value of ttt, how is a point on a parametric curve found?

A

If t=x+13t=\frac{x+1}{3}t=3x+1​, then t2t^2t2 must be replaced by (x+13)2\boxed{\left(\frac{x+1}{3}\right)^2}(3x+1​)2​.

B

The identity used to eliminate ttt from trigonometric parametrisations is sin⁡2t+cos⁡2t=1\boxed{\sin^2t+\cos^2t=1}sin2t+cos2t=1​.

C

Substitute the same value of ttt into both equations.

D

t=x+13t=\frac{x+1}{3}t=3x+1​

Card 1 of 20

1.6.3 Parametric equations of curves (A-level only) Flashcards

  1. A Level
  2. /Maths
  3. /1.6.3 Parametric equations of curves (A-level only)

20 flashcards on AQA A Level Maths 1.6.3 Parametric equations of curves (A-level only): the key formulae, methods and definitions you need to recall for Paper 1, Paper 2 and Paper 3.

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