Maths Extension 1 Year 11 - Module 1 - Lesson 10
Parametric Curves & Sketching
Name
Date
Class
1. Lesson focus
Parametric curves and sketching. Eliminate t, identify the resulting curve and its allowed range, then use increasing t to describe orientation.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. For x=2t and y=t², eliminate t and identify the curve.
3. For x=t and y=t² with −1≤t≤2, eliminate t and give the initial point, terminal point and direction of travel.
3. Challenge
For x=t² and y=t, describe the Cartesian relation and restriction.
4. Multiple choice
For x=t²−1,y=2t, which point is the vertex?
A. (−1,0)
B. (0,−1)
C. (0,0)
D. (1,0)
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. Given x=t²−1 and y=2t, eliminate t.
t=y/2, hence x=y²/4−1, or y²=4(x+1); the vertex is (−1,0).
2. For x=2t and y=t², eliminate t and identify the curve.
t=x/2, so y=x²/4: an upward-opening parabola with vertex (0,0).
3. For x=t and y=t² with −1≤t≤2, eliminate t and give the initial point, terminal point and direction of travel.
y=x² with −1≤x≤2. As t increases from −1 to 2, the point moves from (−1,1), through (0,0), to (2,4); this is the direction of travel.
Challenge. For x=t² and y=t, describe the Cartesian relation and restriction.
x=y² with x≥0; it is a right-opening parabola and the parametrisation selects y=t.
Multiple choice. For x=t²−1,y=2t, which point is the vertex?
Correct answer: A. At t=0, x=−1 and y=0; equivalently y²=4(x+1) has vertex (−1,0).
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