| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring systematic algebraic manipulation (substituting line into parabola equation), discriminant analysis for tangency conditions, and solving a quadratic in m. While methodical, it demands careful algebra across multiple steps and understanding that tangency corresponds to discriminant = 0. The 'no differentiation' constraint requires the classical geometric approach, making it moderately challenging but still a standard FP1 technique. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Equation is \(y - 4 = m(x - 3)\) | M1A1 | 2 marks |
| (b) Elimination of \(x\); \(4y - 16 = m(y^2 - 12)\); Hence result | M1, A1, A1 | 3 marks |
| (c) Discriminant equated to zero; \((3m - 1)(m - 1) = 0\); Tangents \(y = x + 1, y = \frac{1}{3}x + 3\) | M1, m1A1, A1A1 | 5 marks |
| (d) \(m = 1 \Rightarrow y^2 - 4y + 4 = 0\) so point of contact is \((1, 2)\); \(m = \frac{1}{3} \Rightarrow \frac{1}{3}y^2 - 4y + 12 = 0\) so point of contact is \((9, 6)\) | M1, A1, M1, A1 | 4 marks |
**(a)** Equation is $y - 4 = m(x - 3)$ | M1A1 | 2 marks | OE; M1A0 if one small error
**(b)** Elimination of $x$; $4y - 16 = m(y^2 - 12)$; Hence result | M1, A1, A1 | 3 marks | OE (no fractions); convincingly shown (AG)
**(c)** Discriminant equated to zero; $(3m - 1)(m - 1) = 0$; Tangents $y = x + 1, y = \frac{1}{3}x + 3$ | M1, m1A1, A1A1 | 5 marks | OE; m1 for attempt at solving; A1A1; OE
**(d)** $m = 1 \Rightarrow y^2 - 4y + 4 = 0$ so point of contact is $(1, 2)$; $m = \frac{1}{3} \Rightarrow \frac{1}{3}y^2 - 4y + 12 = 0$ so point of contact is $(9, 6)$ | M1, A1, M1, A1 | 4 marks | OE; $m = 1$ needed for this; OE; $m = \frac{1}{3}$ needed for this
**Total: 14 marks**
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**GRAND TOTAL: 75 marks**9 The diagram shows the parabola $y ^ { 2 } = 4 x$ and the point $A$ with coordinates $( 3,4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{504b79bf-1bcc-4fa7-a7a0-689c21a8b03a-05_732_657_370_689}
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the straight line having gradient $m$ and passing through the point $A ( 3,4 )$.
\item Show that, if this straight line intersects the parabola, then the $y$-coordinates of the points of intersection satisfy the equation
$$m y ^ { 2 } - 4 y + ( 16 - 12 m ) = 0$$
\item By considering the discriminant of the equation in part (b), find the equations of the two tangents to the parabola which pass through $A$.\\
(No credit will be given for solutions based on differentiation.)
\item Find the coordinates of the points at which these tangents touch the parabola.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2008 Q9 [14]}}