| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Challenging +1.2 This is a multi-part question involving standard ellipse manipulation and tangency conditions. Parts (a)-(c) are routine: finding intercepts, applying translation, and substituting to get a quadratic. Part (d) requires using the discriminant condition (b²-4ac=0) for tangency, which is a standard FP1 technique explicitly prompted by the question structure. While it involves several steps and algebraic manipulation, it follows a well-established method without requiring novel insight. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | \((\pm\sqrt{5}, 0), (0, \pm 2)\) | B2,1 |
| 8(b) | \(\frac{(x-p)^2}{5} + \frac{y^2}{4} = 1\) | M1 |
| 8(b) | A1 | |
| 8(c) | \(\frac{(x-p)^2}{5} + \frac{(x+4)^2}{1}\) | M1 |
| 8(c) | \(4(x - p)^2 + 5(x + 4)^2 = 4 \times 5\) | |
| 8(c) | \(4(x^2 - 2px + p^2) + 5(x^2 + 8x + 16) = 20\) | m1 |
| 8(c) | \(4x^2 - 8px + 4p^2 + 5x^2 + 40x + 80 = 20\) | |
| 8(c) | \(9x^2 - (8p - 40)x + 4p^2 + 60 = 0\) | A1 |
| 8(d) | Discriminant is \((8p - 40)^2 - 4(9)(4p^2 + 60)\) | B1 |
| 8(d) | For tangency \((8p - 40)^2 - 4(9)(4p^2 + 60) = 0\) | M1 |
| 8(d) | \(p^2 + 8p + 7 = 0\) | A1 |
| 8(d) | \(\{(p + 1)(p + 7) = 0\} \Rightarrow p = -1, p = -7(*)\) | B1 |
| 8(d) | \(p = -1: 9x^2 + 48x + 64 (= 0) \Rightarrow x = -\frac{8}{3}\) | M1 |
| 8(d) | A1 | |
| 8(d) | \(p = -7: 9x^2 + 96x + 256 (= 0) \Rightarrow x = -\frac{16}{3}\) | M1 |
| 8(d) | A1 | |
| 8(d) | \(x = -\frac{8}{3}, y = \frac{4}{3}; \quad x = -\frac{16}{3}, y = -\frac{4}{3}\) | A1 |
8(a) | $(\pm\sqrt{5}, 0), (0, \pm 2)$ | B2,1 | If not B2, award B1 if either at least two of these 4 correct pts or if $^* x = \pm\sqrt{5}$ and $y = \pm 2*$
8(b) | $\frac{(x-p)^2}{5} + \frac{y^2}{4} = 1$ | M1 | Replacing x by either $x + p$ or $x - p$ and keeping y unchanged or as $y \pm 0$
8(b) | | A1 | ACF
8(c) | $\frac{(x-p)^2}{5} + \frac{(x+4)^2}{1}$ | M1 | Substitution into c's (b) eqn of $y = x+4$ to eliminate y
8(c) | $4(x - p)^2 + 5(x + 4)^2 = 4 \times 5$ | | Denominators 5 and 4 cleared in a correct manner and at least either a correct expansion of $(x + p)^2$ or a correct expansion of $(x + 4)^2$
8(c) | $4(x^2 - 2px + p^2) + 5(x^2 + 8x + 16) = 20$ | m1 |
8(c) | $4x^2 - 8px + 4p^2 + 5x^2 + 40x + 80 = 20$ | |
8(c) | $9x^2 - (8p - 40)x + 4p^2 + 60 = 0$ | A1 | CSO No errors in any line of working. AG. Must see brackets correctly removed and all terms involving x, p correctly rearranged to same side before the printed answer is stated. Must have '= 0' although brackets around $4p^2 + 60$ may be omitted
8(d) | Discriminant is $(8p - 40)^2 - 4(9)(4p^2 + 60)$ | B1 | OE Must be isolated, not just within the quadratic formula
8(d) | For tangency $(8p - 40)^2 - 4(9)(4p^2 + 60) = 0$ | M1 | OE Equating c's discriminant to zero before obtaining any values for p
8(d) | $p^2 + 8p + 7 = 0$ | A1 | ACF with like terms collected
8(d) | $\{(p + 1)(p + 7) = 0\} \Rightarrow p = -1, p = -7(*)$ | B1 | Correct values −1, −7 for p. Substitutes at least one of c's two values for p either into the given quadratic in (c) or into $\frac{8p - 40}{18}$
8(d) | $p = -1: 9x^2 + 48x + 64 (= 0) \Rightarrow x = -\frac{8}{3}$ | M1 | OE or into $\frac{8p - 40}{18}$
8(d) | | A1 | $x = -\frac{8}{3}$ OE as only root from the quadratic or from $\frac{8p - 40}{18}$. Apply FiW if (*) is B0
8(d) | $p = -7: 9x^2 + 96x + 256 (= 0) \Rightarrow x = -\frac{16}{3}$ | M1 |
8(d) | | A1 | $x = -\frac{16}{3}$ OE as only root from the quadratic or from $\frac{8p - 40}{18}$. Apply FiW if (*) is B0
8(d) | $x = -\frac{8}{3}, y = \frac{4}{3}; \quad x = -\frac{16}{3}, y = -\frac{4}{3}$ | A1 | CSO Previous 7 marks must have been awarded and coordinates of both points need to be correct and exact but accept in either format. $(-\frac{8}{3}, \frac{4}{3})$ $(-\frac{16}{3}, -\frac{4}{3})$8 The diagram shows the ellipse $E$ with equation
$$\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 4 } = 1$$
and the straight line $L$ with equation
$$y = x + 4$$
\includegraphics[max width=\textwidth, alt={}, center]{9f8cd5ed-f5cf-4cf6-8c92-9fd0819238ca-5_675_1120_708_468}
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of the points where the ellipse $E$ intersects the coordinate axes.
\item The ellipse $E$ is translated by the vector $\left[ \begin{array} { c } p \\ 0 \end{array} \right]$, where $p$ is a constant. Write down the equation of the translated ellipse.
\item Show that, if the translated ellipse intersects the line $L$, the $x$-coordinates of the points of intersection must satisfy the equation
$$9 x ^ { 2 } - ( 8 p - 40 ) x + \left( 4 p ^ { 2 } + 60 \right) = 0$$
\item Given that the line $L$ is a tangent to the translated ellipse, find the coordinates of the two possible points of contact.\\
(No credit will be given for solutions based on differentiation.)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q8 [15]}}