OCR MEI C4 2010 June — Question 3 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeConvert to Cartesian (sin/cos identities)
DifficultyModerate -0.3 This is a straightforward parametric-to-Cartesian conversion requiring standard double-angle identities (cos 2θ = 1 - 2sin²θ and sin θ cos θ = ½sin 2θ). The algebraic manipulation is direct with no tricky steps, and sketching an ellipse is routine. Slightly easier than average due to the standard nature of the identities and techniques involved.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1
3 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 3:
AnswerMarks Guidance
Answer/WorkingMarks 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
\(\Rightarrow x^2 + (2y)^2 = 1 \Rightarrow x^2 + 4y^2 = 1\)E1
*Or:* \(x^2 + 4y^2 = (\cos 2\theta)^2 + 4(\sin\theta\cos\theta)^2\)M1 substitution
\(= \cos^2 2\theta + \sin^2 2\theta = 1\)M1, E1 use of \(\sin 2\theta\)
*Or:* \(\cos^2\theta = (x+1)/2\), \(\sin^2\theta = (1-x)/2\)M1 for both
\(y^2 = \sin^2\theta\cos^2\theta = \left(\frac{1-x}{2}\right)\left(\frac{x+1}{2}\right)\)M1
\(y^2 = (1-x^2)/4\), \(x^2 + 4y^2 = 1\)E1
Ellipse sketchM1 ellipse
Correct intercepts at \((\pm 1, 0)\) and \((0, \pm\frac{1}{2})\)A1 [5] correct intercepts 
# Question 3:

| Answer/Working | 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 | |
| $\Rightarrow x^2 + (2y)^2 = 1 \Rightarrow x^2 + 4y^2 = 1$ | E1 | |
| *Or:* $x^2 + 4y^2 = (\cos 2\theta)^2 + 4(\sin\theta\cos\theta)^2$ | M1 | substitution |
| $= \cos^2 2\theta + \sin^2 2\theta = 1$ | M1, E1 | use of $\sin 2\theta$ |
| *Or:* $\cos^2\theta = (x+1)/2$, $\sin^2\theta = (1-x)/2$ | M1 | for both |
| $y^2 = \sin^2\theta\cos^2\theta = \left(\frac{1-x}{2}\right)\left(\frac{x+1}{2}\right)$ | M1 | |
| $y^2 = (1-x^2)/4$, $x^2 + 4y^2 = 1$ | E1 | |
| Ellipse sketch | M1 | ellipse |
| Correct intercepts at $(\pm 1, 0)$ and $(0, \pm\frac{1}{2})$ | A1 [5] | correct intercepts |

---
3 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 2010 Q3 [5]}}
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