OCR MEI C4 — Question 5 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeConvert to Cartesian (polynomial/rational)
DifficultyModerate -0.3 This is a straightforward parametric-to-Cartesian conversion requiring algebraic manipulation. Students rearrange x = 1/(1+t) to find t = (1-x)/x, then substitute into y and simplify. While it involves rational expressions requiring careful algebra, it's a standard C4 technique with no conceptual difficulty or novel insight required—slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian
5 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(x = \frac{1}{1+t} \Rightarrow 1+t = \frac{1}{x} \Rightarrow t = \frac{1}{x} - 1\)M1, A1 attempt to solve for \(t\)
\(y = \frac{1-t}{1+2t} = \frac{1 - \frac{1}{x} + 1}{1 + \frac{2}{x} - 2}\)M1 substituting for \(t\) in terms of \(x\)
\(= \frac{2 - \frac{1}{x}}{\frac{2}{x} - 1} = \frac{2x-1}{2-x}\)M1, A1 [5] clearing subsidiary fractions 
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \frac{1}{1+t} \Rightarrow 1+t = \frac{1}{x} \Rightarrow t = \frac{1}{x} - 1$ | M1, A1 | attempt to solve for $t$ |
| $y = \frac{1-t}{1+2t} = \frac{1 - \frac{1}{x} + 1}{1 + \frac{2}{x} - 2}$ | M1 | substituting for $t$ in terms of $x$ |
| $= \frac{2 - \frac{1}{x}}{\frac{2}{x} - 1} = \frac{2x-1}{2-x}$ | M1, A1 [5] | clearing subsidiary fractions |

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5 A curve is defined parametrically by the equations

$$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$

Find $t$ in terms of $x$. Hence find the cartesian equation of the curve, giving your answer as simply as possible.

\hfill \mbox{\textit{OCR MEI C4  Q5 [5]}}
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Q2 18 Q3 8 Q4 7 Q5 5 Q6 8