| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2011 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Simple rational function analysis |
| Difficulty | Challenging +1.8 This AEA question requires finding a normal equation (standard calculus), then solving a geometric constraint involving a circle touching three curve branches. Part (c) involves algebraic manipulation to establish identities and solve for the circle parameters. While requiring multiple techniques and careful algebra, the steps are guided and the problem structure is clear, making it challenging but accessible for strong A-level students. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations |
% Figure 2 shows curve with vertical asymptotes at x = -2 and x = 2, horizontal asymptote at y = 1, with U-shaped region between asymptotes
\includegraphics{figure_2}
Figure 2
Figure 2 shows a sketch of the curve $C$ with equation $y = \frac{x^2 - 2}{x^2 - 4}$ and $x \neq \pm 2$.
The curve cuts the $y$-axis at $U$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of the point $U$.
[1]
\end{enumerate}
The point $P$ with $x$-coordinate $a$ ($a \neq 0$) lies on $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the normal to $C$ at $P$ cuts the $y$-axis at the point
$$\left(0, \frac{a^2 - 2}{a^2 - 4} - \frac{(a^2 - 4)^2}{4}\right)$$
[6]
\end{enumerate}
The circle $E$, with centre on the $y$-axis, touches all three branches of $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$\frac{a^2}{2(a^2-4)} - \frac{(a^2-4)^2}{4} = a^2 + \frac{(a^2-4)^4}{16}$$
\item Hence, show that
$$(a^2 - 4)^2 = 1$$
\item Find the centre and radius of $E$.
\end{enumerate}
[10]
\end{enumerate}
[Total 17 marks]
\hfill \mbox{\textit{Edexcel AEA 2011 Q5 [17]}}