| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find tangent equation |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: substituting y=0 to find x-axis intersection, eliminating the parameter (simple since y=t-3 gives t directly), and finding a tangent using dy/dx = (dy/dt)/(dx/dt). All three parts are routine C4 exercises with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solve \(0 = t-3\) & subst into \(x = t^2 - 6t + 4\) | M1 | |
| Obtain \(x = -5\) | A1 (2) | \((-5,0)\) need not be quoted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to eliminate \(t\) | M1 | |
| Simplify to \(x = y^2 - 5\) ISW | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to find \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) from cartes or para form | M1 | Award anywhere in question |
| Obtain \(\frac{dy}{dx} = \frac{1}{2t-6}\) or \(\frac{1}{2y}\) or \((-)\frac{1}{2}(x+5)^{-\frac{1}{2}}\) | A1 | |
| If \(t=2\), \(x=-4\) and \(y=-1\) | B1 | Awarded anywhere in (iii) |
| Using their num \((x,y)\) & their num \(\frac{dy}{dx}\), find tgt eqn | M1 | |
| \(x + 2y + 6 = 0\) AEF(without fractions) ISW | A1 (5) | |
| Total: 9 |
# Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve $0 = t-3$ & subst into $x = t^2 - 6t + 4$ | M1 | |
| Obtain $x = -5$ | A1 **(2)** | $(-5,0)$ need not be quoted |
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# Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to eliminate $t$ | M1 | |
| Simplify to $x = y^2 - 5$ ISW | A1 **(2)** | |
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# Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to find $\frac{dy}{dx}$ or $\frac{dx}{dy}$ from cartes or para form | M1 | Award anywhere in question |
| Obtain $\frac{dy}{dx} = \frac{1}{2t-6}$ or $\frac{1}{2y}$ or $(-)\frac{1}{2}(x+5)^{-\frac{1}{2}}$ | A1 | |
| If $t=2$, $x=-4$ and $y=-1$ | B1 | Awarded anywhere in (iii) |
| Using their num $(x,y)$ & their num $\frac{dy}{dx}$, find tgt eqn | M1 | |
| $x + 2y + 6 = 0$ AEF(without fractions) ISW | A1 **(5)** | |
| **Total: 9** | | |
---6 A curve has parametric equations
$$x = t ^ { 2 } - 6 t + 4 , \quad y = t - 3 .$$
Find\\
(i) the coordinates of the point where the curve meets the $x$-axis,\\
(ii) the equation of the curve in cartesian form, giving your answer in a simple form without brackets,\\
(iii) the equation of the tangent to the curve at the point where $t = 2$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C4 2009 Q6 [9]}}