OCR MEI C4 — Question 8 4 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeConvert to Cartesian (polynomial/rational)
DifficultyModerate -0.8 This is a straightforward parametric-to-Cartesian conversion requiring algebraic manipulation. Students rearrange x = 1/t - 1 to find t = 1/(x+1), substitute into y, and simplify—a routine C4 technique with no conceptual challenges or novel problem-solving required.
Spec1.03g Parametric equations: of curves and conversion to cartesian
8 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is \(y = \frac { 3 + 2 x } { 2 + x }\).
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x = \frac{1}{t} - 1 \Rightarrow \frac{1}{t} = x+1\)M1 Solving for \(t\) in terms of \(x\) or \(y\)
\(t = \frac{1}{x+1}\)A1
\(y = \dfrac{2 + \frac{1}{x+1}}{1 + \frac{1}{x+1}} = \frac{2x+2+1}{x+1+1} = \frac{2x+3}{x+2}\)M1 E1 Subst their \(t\) (must include a fraction), clearing subsidiary fractions/changing subject oe, www
[4]
*or* \(\frac{3+2x}{2+x} = \dfrac{3+\frac{2-2t}{t}}{2+\frac{1-t}{t}} = \frac{3t+2-2t}{2t+1-t} = \frac{t+2}{t+1} = y\)M1 A1 M1 E1 substituting for \(x\) or \(y\) in terms of \(t\); clearing subsidiary fractions/changing subject
[4] 
## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{1}{t} - 1 \Rightarrow \frac{1}{t} = x+1$ | M1 | Solving for $t$ in terms of $x$ or $y$ |
| $t = \frac{1}{x+1}$ | A1 | |
| $y = \dfrac{2 + \frac{1}{x+1}}{1 + \frac{1}{x+1}} = \frac{2x+2+1}{x+1+1} = \frac{2x+3}{x+2}$ | M1 E1 | Subst their $t$ (must include a fraction), clearing subsidiary fractions/changing subject oe, www |
| **[4]** | | |
| ***or*** $\frac{3+2x}{2+x} = \dfrac{3+\frac{2-2t}{t}}{2+\frac{1-t}{t}} = \frac{3t+2-2t}{2t+1-t} = \frac{t+2}{t+1} = y$ | M1 A1 M1 E1 | substituting for $x$ or $y$ in terms of $t$; clearing subsidiary fractions/changing subject |
| **[4]** | | |
8 A curve is defined by parametric equations

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

Show that the cartesian equation of the curve is $y = \frac { 3 + 2 x } { 2 + x }$.

\hfill \mbox{\textit{OCR MEI C4  Q8 [4]}}
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Q1 8 Q2 18 Q3 7 Q4 5 Q5 18 Q6 7 Q7 5 Q8 4