OCR MEI C4 — Question 4 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeConvert to Cartesian (sin/cos identities)
DifficultyStandard +0.3 This is a straightforward parametric-to-Cartesian conversion using standard double-angle identities (sin 2θ = 2sin θ cos θ, cos 2θ = cos²θ - sin²θ). The 'show that' format makes it easier as students know the target. The sketch requires recognizing an ellipse, which is routine. Slightly above trivial due to the identity manipulation, but below average difficulty overall.
Spec1.02n Sketch curves: simple equations including polynomials1.03g Parametric equations: of curves and conversion to cartesian
4 The parametric equations of a curve are $$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$ Show that the cartesian equation of the curve is \(x ^ { 2 } + 4 y ^ { 2 } = 1\).
Sketch the curve.
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(y = \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta\)M1 use of \(\sin 2\theta\)
\(x = \cos 2\theta\); \(\sin^2 2\theta + \cos^2 2\theta = 1\)M1, E1
\(\Rightarrow x^2 + (2y)^2 = 1 \Rightarrow x^2 + 4y^2 = 1\) ✓M1, M1, E1 substitution; use of \(\sin 2\theta\)
*or* \(x^2 + 4y^2 = (\cos 2\theta)^2 + 4(\sin\theta\cos\theta)^2 = \cos^2 2\theta + \sin^2 2\theta = 1\) ✓
*or via* \(\cos 2\theta = 2\cos^2\theta - 1\): \(\cos^2\theta = (x+1)/2\); \(\cos 2\theta = 1 - 2\sin^2\theta\): \(\sin^2\theta = (1-x)/2\)M1 for both
\(y^2 = \sin^2\theta\cos^2\theta = \left(\dfrac{1-x}{2}\right)\left(\dfrac{x+1}{2}\right)\); \(y^2 = (1-x^2)/4\); \(x^2 + 4y^2 = 1\) ✓M1, E1
*or* \(x = \cos^4\theta - 2\sin^2\theta\cos^2\theta + \sin^4\theta\); \(y^2 = \sin^2\theta\cos^2\theta\)M1 correct use of double angle formulae
\(x^2 + 4y^2 = (\cos^2\theta + \sin^2\theta)^2 = 1\) ✓M1, E1 correct squaring and use of \(\sin^2\theta + \cos^2\theta = 1\)
Sketch: ellipse with correct intercepts (±1 on \(x\)-axis, \(\pm\frac{1}{2}\) on \(y\)-axis)M1, A1 ellipse; correct intercepts
[5]
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta$ | M1 | use of $\sin 2\theta$ |
| $x = \cos 2\theta$; $\sin^2 2\theta + \cos^2 2\theta = 1$ | M1, E1 | |
| $\Rightarrow x^2 + (2y)^2 = 1 \Rightarrow x^2 + 4y^2 = 1$ ✓ | M1, M1, E1 | substitution; use of $\sin 2\theta$ |
| *or* $x^2 + 4y^2 = (\cos 2\theta)^2 + 4(\sin\theta\cos\theta)^2 = \cos^2 2\theta + \sin^2 2\theta = 1$ ✓ | | |
| *or via* $\cos 2\theta = 2\cos^2\theta - 1$: $\cos^2\theta = (x+1)/2$; $\cos 2\theta = 1 - 2\sin^2\theta$: $\sin^2\theta = (1-x)/2$ | M1 | for both |
| $y^2 = \sin^2\theta\cos^2\theta = \left(\dfrac{1-x}{2}\right)\left(\dfrac{x+1}{2}\right)$; $y^2 = (1-x^2)/4$; $x^2 + 4y^2 = 1$ ✓ | M1, E1 | |
| *or* $x = \cos^4\theta - 2\sin^2\theta\cos^2\theta + \sin^4\theta$; $y^2 = \sin^2\theta\cos^2\theta$ | M1 | correct use of double angle formulae |
| $x^2 + 4y^2 = (\cos^2\theta + \sin^2\theta)^2 = 1$ ✓ | M1, E1 | correct squaring and use of $\sin^2\theta + \cos^2\theta = 1$ |
| Sketch: ellipse with correct intercepts (±1 on $x$-axis, $\pm\frac{1}{2}$ on $y$-axis) | M1, A1 | ellipse; correct intercepts |

**[5]**
4 The parametric equations of a curve are

$$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$

Show that the cartesian equation of the curve is $x ^ { 2 } + 4 y ^ { 2 } = 1$.\\
Sketch the curve.

\hfill \mbox{\textit{OCR MEI C4  Q4 [5]}}
This paper (8 questions)
View full paper
Q1 8 Q2 18 Q3 7 Q4 5 Q5 18 Q6 7 Q7 5 Q8 4