For a fixed value of ttt, how is a point on a parametric curve found?
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.
The identity used to eliminate ttt from trigonometric parametrisations is sin2t+cos2t=1\boxed{\sin^2t+\cos^2t=1}sin2t+cos2t=1.
Substitute the same value of ttt into both equations.
t=x+13t=\frac{x+1}{3}t=3x+1
1.6.3 Parametric equations of curves (A-level only) Flashcards
Flashcards for AQA A Level Maths 1.6.3 Parametric equations of curves (A-level only), covering the key formulae, methods and definitions you need to recall for Paper 1, Paper 2 and Paper 3. 20 cards, matched to the AQA A Level Maths (7357) specification. Recall questions account for roughly 50% of marks at A Level Maths, so these target the marks you can secure before the paper starts.