| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Normal passes through specific point verification |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: substituting coordinates to find parameter value, finding dy/dx using the chain rule, determining the normal equation, and eliminating the parameter to get Cartesian form. All steps are routine C4 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 1\) \(\therefore -1 + 4\cos\theta = 1\), \(\cos\theta = \frac{1}{2}\), \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\) | M1 | |
| \(y > 0\) \(\therefore \sin\theta > 0\) \(\therefore \theta = \frac{\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = -4\sin\theta\), \(\quad \frac{dy}{d\theta} = 2\sqrt{2}\cos\theta\) | M1 | |
| \(\therefore \frac{dy}{dx} = \frac{2\sqrt{2}\cos\theta}{-4\sin\theta}\) | M1 A1 | |
| at \(P\), \(\text{grad} = -\frac{2\sqrt{2} \times \frac{1}{2}}{4 \times \frac{\sqrt{3}}{2}} = -\frac{\sqrt{2}}{2\sqrt{3}}\) | M1 | |
| grad of normal \(= \frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{6}\) | A1 | |
| \(\therefore y - \sqrt{6} = \sqrt{6}(x-1)\) | M1 | |
| \(y = \sqrt{6}\,x\), \(\quad\) when \(x = 0\), \(y = 0\) \(\therefore\) passes through origin | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \frac{x+1}{4}\), \(\sin\theta = \frac{y}{2\sqrt{2}}\) | M1 | |
| \(\therefore \frac{(x+1)^2}{16} + \frac{y^2}{8} = 1\) | M1 A1 | (12) |
| Total | (72) |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 1$ $\therefore -1 + 4\cos\theta = 1$, $\cos\theta = \frac{1}{2}$, $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$ | M1 | |
| $y > 0$ $\therefore \sin\theta > 0$ $\therefore \theta = \frac{\pi}{3}$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = -4\sin\theta$, $\quad \frac{dy}{d\theta} = 2\sqrt{2}\cos\theta$ | M1 | |
| $\therefore \frac{dy}{dx} = \frac{2\sqrt{2}\cos\theta}{-4\sin\theta}$ | M1 A1 | |
| at $P$, $\text{grad} = -\frac{2\sqrt{2} \times \frac{1}{2}}{4 \times \frac{\sqrt{3}}{2}} = -\frac{\sqrt{2}}{2\sqrt{3}}$ | M1 | |
| grad of normal $= \frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{6}$ | A1 | |
| $\therefore y - \sqrt{6} = \sqrt{6}(x-1)$ | M1 | |
| $y = \sqrt{6}\,x$, $\quad$ when $x = 0$, $y = 0$ $\therefore$ passes through origin | A1 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \frac{x+1}{4}$, $\sin\theta = \frac{y}{2\sqrt{2}}$ | M1 | |
| $\therefore \frac{(x+1)^2}{16} + \frac{y^2}{8} = 1$ | M1 A1 | **(12)** |
| | Total | **(72)** |8.\\
\includegraphics[max width=\textwidth, alt={}, center]{85427816-dcf1-49af-8d68-f4e88fc7d8f1-3_497_784_246_461}
The diagram shows the curve with parametric equations
$$x = - 1 + 4 \cos \theta , \quad y = 2 \sqrt { 2 } \sin \theta , \quad 0 \leq \theta < 2 \pi$$
The point $P$ on the curve has coordinates $( 1 , \sqrt { 6 } )$.\\
(i) Find the value of $\theta$ at $P$.\\
(ii) Show that the normal to the curve at $P$ passes through the origin.\\
(iii) Find a cartesian equation for the curve.
\hfill \mbox{\textit{OCR C4 Q8 [12]}}