| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.2 This is a structured multi-part question on ellipse tangents requiring implicit differentiation (standard FP3 technique), circle tangent equation (routine), solving simultaneous equations, and finding a locus by eliminating a parameter. While it involves several steps and Further Maths content, each part follows predictable methods with clear guidance, making it moderately above average difficulty but not requiring significant novel insight. |
| Spec | 1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0\) and so \(\frac{dy}{dx} = -\frac{xb^2}{ya^2} = -\frac{b\cos\theta}{a\sin\theta}\) | M1 A1 | Finding gradient in terms of \(\theta\), must use calculus; cao |
| \(y - b\sin\theta = -\frac{b\cos\theta}{a\sin\theta}(x - a\cos\theta)\) | M1 | Finding equation of tangent |
| Uses \(\cos^2\theta + \sin^2\theta = 1\) to give \(\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1\) | A1cso | cso. Need to get \(\cos^2\theta + \sin^2\theta\) on the same side |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of circle is \(-\frac{\cos\theta}{\sin\theta}\), equation of tangent: \(y - a\sin\theta = -\frac{\cos\theta}{\sin\theta}(x - a\cos\theta)\) or set \(a = b\) in previous answer | M1 | Finding gradient and equation of tangent, or setting \(a = b\) |
| So \(y\sin\theta + x\cos\theta = a\) | A1 | cao, need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eliminate \(x\) or \(y\) to give \(y\sin\theta\left(\frac{a}{b}-1\right) = 0\) or \(x\cos\theta\left(\frac{b}{a}-1\right) = b - a\) | M1 | As scheme |
| \(l_1\) and \(l_2\) meet at \(\left(\dfrac{a}{\cos\theta}, 0\right)\) | A1, B1 | \(x = \frac{a}{\cos\theta}\), need not be simplified; \(y = 0\), need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The locus of \(R\) is part of the line \(y = 0\), such that \(x \geq a\) and \(x \leq -a\). Or clearly labelled sketch. Accept "real axis" | B1, B1 | Identifying locus as \(y=0\) or real/\(x\) axis; identifies correct parts of \(y=0\), condone use of strict inequalities |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0$ and so $\frac{dy}{dx} = -\frac{xb^2}{ya^2} = -\frac{b\cos\theta}{a\sin\theta}$ | M1 A1 | Finding gradient in terms of $\theta$, must use calculus; cao |
| $y - b\sin\theta = -\frac{b\cos\theta}{a\sin\theta}(x - a\cos\theta)$ | M1 | Finding equation of tangent |
| Uses $\cos^2\theta + \sin^2\theta = 1$ to give $\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$ | A1cso | cso. Need to get $\cos^2\theta + \sin^2\theta$ on the same side |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of circle is $-\frac{\cos\theta}{\sin\theta}$, equation of tangent: $y - a\sin\theta = -\frac{\cos\theta}{\sin\theta}(x - a\cos\theta)$ or set $a = b$ in previous answer | M1 | Finding gradient and equation of tangent, or setting $a = b$ |
| So $y\sin\theta + x\cos\theta = a$ | A1 | cao, need not be simplified |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eliminate $x$ or $y$ to give $y\sin\theta\left(\frac{a}{b}-1\right) = 0$ or $x\cos\theta\left(\frac{b}{a}-1\right) = b - a$ | M1 | As scheme |
| $l_1$ and $l_2$ meet at $\left(\dfrac{a}{\cos\theta}, 0\right)$ | A1, B1 | $x = \frac{a}{\cos\theta}$, need not be simplified; $y = 0$, need not be simplified |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The locus of $R$ is part of the line $y = 0$, such that $x \geq a$ and $x \leq -a$. Or clearly labelled sketch. Accept "real axis" | B1, B1 | Identifying locus as $y=0$ or real/$x$ axis; identifies correct parts of $y=0$, condone use of strict inequalities |\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
The line $l _ { 1 }$ is a tangent to $E$ at the point $P ( a \cos \theta , b \sin \theta )$.\\
(a) Using calculus, show that an equation for $l _ { 1 }$ is
$$\frac { x \cos \theta } { a } + \frac { y \sin \theta } { b } = 1$$
The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$$
The line $l _ { 2 }$ is a tangent to $C$ at the point $Q ( a \cos \theta , a \sin \theta )$.\\
(b) Find an equation for the line $l _ { 2 }$.
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $R$,\\
(c) find, in terms of $a , b$ and $\theta$, the coordinates of $R$.\\
(d) Find the locus of $R$, as $\theta$ varies.\\
\hfill \mbox{\textit{Edexcel FP3 2012 Q6 [11]}}