| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 Part (a) is routine conversion from parametric to Cartesian form using cos²θ + sin²θ = 1, requiring only algebraic manipulation. Part (b) requires finding dy/dx using the chain rule for parametric equations, forming the tangent equation, and finding the x-intercept—all standard A-level techniques with no novel insight needed. The multi-step nature and 'show detailed reasoning' requirement add slight difficulty, but this remains a straightforward textbook-style question. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-3)^2 + (y+4)^2 = 4\cos^2\theta + 4\sin^2\theta\) | M1 | or \(\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+4}{2}\right)^2 = \cos^2\theta + \sin^2\theta\); Condone sign errors or one arith slip or missing brackets for M1 |
| \(\Rightarrow (x-3)^2 + (y+4)^2 = 4\) oe ISW | A1 | or \(\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+4}{2}\right)^2 = 1\) oe; or \(\cos^{-1}\left(\frac{x-3}{2}\right) = \sin^{-1}\left(\frac{y+4}{2}\right)\) M1A1; or \(y = -4 + 2\sqrt{1-\left(\frac{x-3}{2}\right)^2}\) M1A1; ISW for all answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Centre \((3, -4)\), radius \(2\) | B1f | ft their (i) if both consistent with (i); But if absolutely correct, not ft: B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | Attempt diff \(x\) & \(y\) wrt \(t\) & find \(\frac{dy}{dt} \div \frac{dx}{dt}\); NB Allow decimals to 2 sf instead of surds throughout, except answer to 3 sf |
| \(= -\frac{1}{2}\cot t\) or \(-\frac{1}{2}\frac{\cos t}{\sin t}\) | A1 | soi |
| \(t = \frac{\pi}{6}\): \(\frac{dy}{dx} = -\frac{1}{2}\cot\frac{\pi}{6}\) oe or \(-\frac{\sqrt{3}}{2}\) oe | M1 | Substitute \(t = \frac{\pi}{6}\) in their \(\frac{dy}{dx}\); Allow sign error |
| Alternative methods for gradient: \(\left(\frac{x}{4}\right)^2 + \left(\frac{y}{2}\right)^2 = 1\), \(\frac{x}{8} + \frac{y}{2}\frac{dy}{dx} = 0\) | M1 | Attempt cartesian eqn & differentiation; \(\frac{d}{dx}(0.5(16-x^2)^{-0.5})\) |
| \(\frac{dy}{dx} = -\frac{x}{4y}\) | A1 | soi; or \(\frac{dy}{dx} = \frac{1}{4}(16-x^2)^{-0.5}(-2x)\) oe |
| \(t = \frac{\pi}{6}\): \(\frac{dy}{dx} = \frac{4\cos(\pi/6)}{8\sin(\pi/6)}\) or \(-\frac{1}{2}\cot\frac{\pi}{6}\) or \(-\frac{\sqrt{3}}{2}\) | M1 | Substitute \(t=\frac{\pi}{6}\) in \(x\) (and \(y\)) & their \(\frac{dy}{dx}\) |
| Eqn of \(L\): \(y - 2\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2}\left(x - 4\cos\frac{\pi}{6}\right)\) oe | M1 | or \(y=-\frac{\sqrt{3}}{2}x+c\) & subst \(\left(4\cos\frac{\pi}{6}, 2\sin\frac{\pi}{6}\right)\); ft their grad (not \(-ve\) reciprocal); Must not involve \(t\) |
| or \(y - 1 = -\frac{\sqrt{3}}{2}(x - 2\sqrt{3})\) oe; or \(y = -\frac{\sqrt{3}}{2}x + 4\) oe | This mark may be implied by next mark | |
| \(0 - 1 = -\frac{\sqrt{3}}{2}x + 3\) oe | M1 | or \(0 = -\frac{\sqrt{3}}{2}x + 4\) oe; Subst \(y=0\) in their line eqn, not involving \(t\) |
| Cuts at \(\left(\frac{8\sqrt{3}}{3}, 0\right)\) oe or \((4.62, 0)\) (3 sf) | A1 | Allow just \(\frac{8\sqrt{3}}{3}\) or \(4.62\) (3 sf); Allow equivalents, eg \(\frac{8}{\sqrt{3}}\) |
# Question 3:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-3)^2 + (y+4)^2 = 4\cos^2\theta + 4\sin^2\theta$ | M1 | or $\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+4}{2}\right)^2 = \cos^2\theta + \sin^2\theta$; Condone sign errors or one arith slip or missing brackets for M1 |
| $\Rightarrow (x-3)^2 + (y+4)^2 = 4$ oe ISW | A1 | or $\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+4}{2}\right)^2 = 1$ oe; or $\cos^{-1}\left(\frac{x-3}{2}\right) = \sin^{-1}\left(\frac{y+4}{2}\right)$ M1A1; or $y = -4 + 2\sqrt{1-\left(\frac{x-3}{2}\right)^2}$ M1A1; **ISW for all answers** |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre $(3, -4)$, radius $2$ | B1f | ft their (i) if both consistent with (i); But if absolutely correct, not ft: B1 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | Attempt diff $x$ & $y$ wrt $t$ & find $\frac{dy}{dt} \div \frac{dx}{dt}$; **NB Allow decimals to 2 sf instead of surds throughout, except answer to 3 sf** |
| $= -\frac{1}{2}\cot t$ or $-\frac{1}{2}\frac{\cos t}{\sin t}$ | A1 | soi |
| $t = \frac{\pi}{6}$: $\frac{dy}{dx} = -\frac{1}{2}\cot\frac{\pi}{6}$ oe or $-\frac{\sqrt{3}}{2}$ oe | M1 | Substitute $t = \frac{\pi}{6}$ in their $\frac{dy}{dx}$; Allow sign error |
| **Alternative methods for gradient:** $\left(\frac{x}{4}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$, $\frac{x}{8} + \frac{y}{2}\frac{dy}{dx} = 0$ | M1 | Attempt cartesian eqn & differentiation; $\frac{d}{dx}(0.5(16-x^2)^{-0.5})$ |
| $\frac{dy}{dx} = -\frac{x}{4y}$ | A1 | soi; or $\frac{dy}{dx} = \frac{1}{4}(16-x^2)^{-0.5}(-2x)$ oe |
| $t = \frac{\pi}{6}$: $\frac{dy}{dx} = \frac{4\cos(\pi/6)}{8\sin(\pi/6)}$ or $-\frac{1}{2}\cot\frac{\pi}{6}$ or $-\frac{\sqrt{3}}{2}$ | M1 | Substitute $t=\frac{\pi}{6}$ in $x$ (and $y$) & their $\frac{dy}{dx}$ |
| Eqn of $L$: $y - 2\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2}\left(x - 4\cos\frac{\pi}{6}\right)$ oe | M1 | or $y=-\frac{\sqrt{3}}{2}x+c$ & subst $\left(4\cos\frac{\pi}{6}, 2\sin\frac{\pi}{6}\right)$; ft their grad (not $-ve$ reciprocal); Must not involve $t$ |
| or $y - 1 = -\frac{\sqrt{3}}{2}(x - 2\sqrt{3})$ oe; or $y = -\frac{\sqrt{3}}{2}x + 4$ oe | | This mark may be implied by next mark |
| $0 - 1 = -\frac{\sqrt{3}}{2}x + 3$ oe | M1 | or $0 = -\frac{\sqrt{3}}{2}x + 4$ oe; Subst $y=0$ in their line eqn, not involving $t$ |
| Cuts at $\left(\frac{8\sqrt{3}}{3}, 0\right)$ oe or $(4.62, 0)$ (3 sf) | A1 | Allow just $\frac{8\sqrt{3}}{3}$ or $4.62$ (3 sf); Allow equivalents, eg $\frac{8}{\sqrt{3}}$ |
---3
\begin{enumerate}[label=(\alph*)]
\item A circle is defined by the parametric equations $x = 3 + 2 \cos \theta , y = - 4 + 2 \sin \theta$.
\begin{enumerate}[label=(\roman*)]
\item Find a cartesian equation of the circle.
\item Write down the centre and radius of the circle.
\end{enumerate}\item In this question you must show detailed reasoning.
The curve $S$ is defined by the parametric equations $x = 4 \cos t , y = 2 \sin t$. The line $L$ is a tangent to $S$ at the point given by $t = \frac { 1 } { 6 } \pi$.
Find where the line $L$ cuts the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2019 Q3 [9]}}