| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Challenging +1.2 This is a structured multi-part question on hyperbolas with clear scaffolding. Part (a) is routine substitution, part (b) is standard line-hyperbola intersection algebra, part (c) is applying the discriminant condition, and part (d) requires finding tangent points using the equal roots condition. While it involves several steps and the restriction against differentiation adds mild complexity, the question guides students through each stage methodically. It's moderately harder than average due to the algebraic manipulation and the tangent-from-external-point concept, but remains a standard FP1 exercise without requiring novel insight. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| 9(a) | \(x = 2, y = 0 \Rightarrow \frac{4}{a^2} - 0 = 1\) so \(a = 2\) | E2,1 |
| Asymps \(\Rightarrow \pm \frac{b}{a} = \pm 2\) so \(b = 2a = 4\) | E2,1 | ditto |
| 4 marks | ||
| 9(b) | Line is \(y - 0 = m(x - 1)\) | B1 |
| Elimination of \(y\) | M1 | |
| \(4x^2 - m^2(x^2 - 2x + 1) = 16\) | A1 | OE (no fractions) |
| So \((m^2 - 4)x^2 - 2m^2x + (m^2 + 16) = 0\) | A1 | 4 marks |
| 9(c) | Discriminant equated to zero | M1 |
| \(4m^4 - 4m^4 - 64m^2 + 16m^2 + 256 = 0\) | A1 | OE |
| \(-3m^2 + 16 = 0\), hence result | A1 | 3 marks |
| 9(d) | \(m^2 = \frac{16}{3} \Rightarrow \frac{4}{3}x^2 - \frac{32}{3}x + \frac{64}{3} = 0\) | M1 |
| \(x^2 - 8x + 16 = 0\), so \(x = 4\) | m1A1 | |
| Method for \(y\)-coordinates | m1 | using \(m = \pm\frac{4}{\sqrt{3}}\) or from equation of hyperbola; dep't on previous m1 |
| \(y = \pm 4\sqrt{3}\) | A1 | 5 marks |
| Total for Q9 | 16 marks |
| Answer | Marks |
|---|---|
| GRAND TOTAL | 75 marks |
9(a) | $x = 2, y = 0 \Rightarrow \frac{4}{a^2} - 0 = 1$ so $a = 2$ | E2,1 | E1 for verif'n or incomplete proof |
| Asymps $\Rightarrow \pm \frac{b}{a} = \pm 2$ so $b = 2a = 4$ | E2,1 | ditto |
| | | 4 marks |
9(b) | Line is $y - 0 = m(x - 1)$ | B1 | OE |
| Elimination of $y$ | M1 |
| $4x^2 - m^2(x^2 - 2x + 1) = 16$ | A1 | OE (no fractions) |
| So $(m^2 - 4)x^2 - 2m^2x + (m^2 + 16) = 0$ | A1 | 4 marks | convincingly shown (AG) |
9(c) | Discriminant equated to zero | M1 |
| $4m^4 - 4m^4 - 64m^2 + 16m^2 + 256 = 0$ | A1 | OE |
| $-3m^2 + 16 = 0$, hence result | A1 | 3 marks | convincingly shown (AG) |
9(d) | $m^2 = \frac{16}{3} \Rightarrow \frac{4}{3}x^2 - \frac{32}{3}x + \frac{64}{3} = 0$ | M1 |
| $x^2 - 8x + 16 = 0$, so $x = 4$ | m1A1 |
| Method for $y$-coordinates | m1 | using $m = \pm\frac{4}{\sqrt{3}}$ or from equation of hyperbola; dep't on previous m1 |
| $y = \pm 4\sqrt{3}$ | A1 | 5 marks |
| **Total for Q9** | **16 marks** |
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| **GRAND TOTAL** | **75 marks** |9 The diagram shows the hyperbola
$$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
and its asymptotes.\\
\includegraphics[max width=\textwidth, alt={}, center]{3c141dcb-4a5e-45ff-9c8e-e06762c03d10-6_798_939_612_555}
The constants $a$ and $b$ are positive integers.\\
The point $A$ on the hyperbola has coordinates ( 2,0 ).\\
The equations of the asymptotes are $y = 2 x$ and $y = - 2 x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 2$ and $b = 4$.
\item The point $P$ has coordinates ( 1,0 ). A straight line passes through $P$ and has gradient $m$. Show that, if this line intersects the hyperbola, the $x$-coordinates of the points of intersection satisfy the equation
$$\left( m ^ { 2 } - 4 \right) x ^ { 2 } - 2 m ^ { 2 } x + \left( m ^ { 2 } + 16 \right) = 0$$
\item Show that this equation has equal roots if $3 m ^ { 2 } = 16$.
\item There are two tangents to the hyperbola which pass through $P$. Find the coordinates of the points at which these tangents touch the hyperbola.\\
(No credit will be given for solutions based on differentiation.)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2010 Q9 [16]}}