| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse tangent/normal equation derivation |
| Difficulty | Challenging +1.2 This is a standard Further Maths ellipse question requiring implicit differentiation to find the normal equation, then algebraic manipulation to find intercepts and optimize area using trigonometry. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are routine for FP3 students: calculus on parametric/implicit forms, finding intercepts, and optimizing with standard trig identities. No novel geometric insight required. |
| Spec | 1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{d\theta}=-a\sin\theta\) and \(\frac{dy}{d\theta}=b\cos\theta\), so \(\frac{dy}{dx}=\frac{b\cos\theta}{-a\sin\theta}\) | M1 A1 | M1: Differentiates both \(x\) and \(y\) and divides correctly. A1: Fully correct derivative |
| Alternative: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{2x}{a^2}+\frac{2yy'}{b^2}=0 \Rightarrow y'=-\frac{b^2x}{a^2y}=-\frac{b^2 a\cos\theta}{a^2 b\sin\theta}\) | M1 A1 | M1: Differentiates implicitly and substitutes for \(x\) and \(y\). A1: \(=-\frac{b\cos\theta}{a\sin\theta}\) |
| Normal has gradient \(\frac{a\sin\theta}{b\cos\theta}\) or \(\frac{a^2y}{b^2x}\) | M1 | Correct perpendicular gradient rule |
| \((y-b\sin\theta)=\frac{a\sin\theta}{b\cos\theta}(x-a\cos\theta)\) | M1 | Correct straight line method using a "changed" gradient which is a function of \(\theta\). If \(y=mx+c\) used, need to find \(c\) for M1 |
| \(ax\sin\theta - by\cos\theta = (a^2-b^2)\sin\theta\cos\theta\) * | A1 | Fully correct completion to printed answer |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=\frac{(a^2-b^2)\cos\theta}{a}\) | B1 | Allow unsimplified |
| \(y=-\frac{(a^2-b^2)\sin\theta}{b}\) | B1 | Allow unsimplified |
| \(\left(=\frac{\frac{1}{2}(a^2-b^2)^2\cos\theta\sin\theta}{ab}\right) = \frac{1}{4}\frac{(a^2-b^2)^2}{ab}\sin 2\theta\) | M1A1 | M1: Area of triangle is \(\frac{1}{2}\)"\(OA\)"\(\times\)"\(OB\)" and uses double angle formula correctly. A1: Correct expression for area (must be positive) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum area when \(\sin 2\theta=1\) so \(\theta=\frac{\pi}{4}\) or \(45°\) | B1 | Correct value for \(\theta\) (may be implied by correct coordinates) |
| So the point \(P\) is at \(\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)\) oe; \(\left(a\cos\frac{\pi}{4}, b\sin\frac{\pi}{4}\right)\) scores B1M1A0 | M1 A1 | M1: Substitutes their value of \(\theta\) where \(0<\theta<\frac{\pi}{2}\) or \(0<\theta<90°\) into their parametric coordinates. A1: Correct exact coordinates |
| (3) | ||
| Total 12 |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{d\theta}=-a\sin\theta$ and $\frac{dy}{d\theta}=b\cos\theta$, so $\frac{dy}{dx}=\frac{b\cos\theta}{-a\sin\theta}$ | M1 A1 | M1: Differentiates both $x$ and $y$ and divides correctly. A1: Fully correct derivative |
| Alternative: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{2x}{a^2}+\frac{2yy'}{b^2}=0 \Rightarrow y'=-\frac{b^2x}{a^2y}=-\frac{b^2 a\cos\theta}{a^2 b\sin\theta}$ | M1 A1 | M1: Differentiates implicitly and substitutes for $x$ and $y$. A1: $=-\frac{b\cos\theta}{a\sin\theta}$ |
| Normal has gradient $\frac{a\sin\theta}{b\cos\theta}$ or $\frac{a^2y}{b^2x}$ | M1 | Correct perpendicular gradient rule |
| $(y-b\sin\theta)=\frac{a\sin\theta}{b\cos\theta}(x-a\cos\theta)$ | M1 | Correct straight line method using a "changed" gradient which is a function of $\theta$. If $y=mx+c$ used, need to find $c$ for M1 |
| $ax\sin\theta - by\cos\theta = (a^2-b^2)\sin\theta\cos\theta$ * | A1 | Fully correct completion to printed answer |
| | | (5) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=\frac{(a^2-b^2)\cos\theta}{a}$ | B1 | Allow unsimplified |
| $y=-\frac{(a^2-b^2)\sin\theta}{b}$ | B1 | Allow unsimplified |
| $\left(=\frac{\frac{1}{2}(a^2-b^2)^2\cos\theta\sin\theta}{ab}\right) = \frac{1}{4}\frac{(a^2-b^2)^2}{ab}\sin 2\theta$ | M1A1 | M1: Area of triangle is $\frac{1}{2}$"$OA$"$\times$"$OB$" and uses double angle formula correctly. A1: Correct expression for area (must be positive) |
| | | (4) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum area when $\sin 2\theta=1$ so $\theta=\frac{\pi}{4}$ or $45°$ | B1 | Correct value for $\theta$ (may be implied by correct coordinates) |
| So the point $P$ is at $\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)$ oe; $\left(a\cos\frac{\pi}{4}, b\sin\frac{\pi}{4}\right)$ scores B1M1A0 | M1 A1 | M1: Substitutes their value of $\theta$ where $0<\theta<\frac{\pi}{2}$ or $0<\theta<90°$ into their parametric coordinates. A1: Correct exact coordinates |
| | | (3) |
| | | **Total 12** |
---\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 , \quad a > b > 0$$
The line $l$ is a normal to $E$ at a point $P ( a \cos \theta , b \sin \theta ) , \quad 0 < \theta < \frac { \pi } { 2 }$\\
(a) Using calculus, show that an equation for $l$ is
$$a x \sin \theta - b y \cos \theta = \left( a ^ { 2 } - b ^ { 2 } \right) \sin \theta \cos \theta$$
The line $l$ meets the $x$-axis at $A$ and the $y$-axis at $B$.\\
(b) Show that the area of the triangle $O A B$, where $O$ is the origin, may be written as $k \sin 2 \theta$, giving the value of the constant $k$ in terms of $a$ and $b$.\\
(c) Find, in terms of $a$ and $b$, the exact coordinates of the point $P$, for which the area of the triangle $O A B$ is a maximum.\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q7 [12]}}