| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| 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 on ellipse tangents requiring substitution, discriminant condition for tangency (b²=4ac), and coordinate finding. While systematic, it demands careful algebraic manipulation across several steps and understanding of the tangency condition, placing it moderately above average difficulty for A-level but standard for FP1. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02q Use intersection points: of graphs to solve equations1.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| \(y=0,\ \dfrac{(x-4)^2}{4}=1;\ (x-4)^2=4\) | M1 | OE. Sub \(y=0\) in eqn of ellipse and either eliminate fraction or take sq root, condoning missing \(\pm\), ie \(\frac{(x-4)}{2} = (\pm)1\) |
| \(\Rightarrow x = 2, 6\ (x_A=2,\ x_B=6)\) | A1 | Both 2 and 6. NMS Mark as B2 or B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{(x-4)^2}{4} + (mx)^2 = 1\) | M1 | Substitute \(y=mx\) to eliminate \(y\) |
| \((x-4)^2 + 4(mx)^2 = 4 \Rightarrow x^2 - 8x + 16 + 4(mx)^2 = 4\) | A1 | Eliminate fractions correctly and expand \((x-4)^2\) correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow (1+4m^2)x^2 - 8x + 12 = 0\) | A1 | CSO AG |
| Answer | Marks | Guidance |
|---|---|---|
| Discriminant \(b^2 - 4ac\): \(\{(-8)^2 - 4(1+4m^2)(12)\}\) | M1 | \(b^2-4ac\) in terms of \(m\), condone one sign or copying error OE |
| For tangency, \((-8)^2 - 4(1+4m^2)(12) = 0\) | A1 | A correct equation with \(m^2\) being the only unknown at any stage. |
| \(192m^2 - 16\ (=0)\) | A1 | OE eg \(12m^2-1(=0)\) OE PI by a correct value for \(m\) condoning wrong sign |
| \((m>0\ \text{so})\ m = \dfrac{1}{\sqrt{12}}\) | A1 | ACF of an exact value for \(m\). eg \(\frac{1}{2\sqrt{3}}, \frac{\sqrt{3}}{6}\). Dep on prev 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(1 + 4\times\left\{\dfrac{1}{\sqrt{12}}\right\}^2\right)x^2 - 8x + 12\ (=0)\) | M1 | Subst value for \(m\) in LHS of eqn (b)(i); ft on c's value of \(m\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 6x + 9 = 0\) | m1 | Valid method to solve a correct quadratic equation; as far as either correct subst into quadratic formula with \(b^2-4ac\) evaluated to 0, or correct factorisation or correct value of \(x\) after \(\frac{4}{3}x^2-8x+12=0\), or better seen; OE, correct use of \(-b/2a\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3\) | A1 | Must see earlier justification |
| Coordinates of \(P\) are \(\left(3,\ \dfrac{3}{\sqrt{12}}\right)\) | A1 | Correct coordinates with the \(y\)-coord in any correct exact form eg \(\frac{\sqrt{3}}{2}\). NMS SC1 for \(\left(3, \frac{3}{\sqrt{12}}\right)\) |
## Question 9:
### Part (a):
$y=0,\ \dfrac{(x-4)^2}{4}=1;\ (x-4)^2=4$ | M1 | OE. Sub $y=0$ in eqn of ellipse and either eliminate fraction or take sq root, condoning missing $\pm$, ie $\frac{(x-4)}{2} = (\pm)1$
$\Rightarrow x = 2, 6\ (x_A=2,\ x_B=6)$ | A1 | Both 2 and 6. NMS Mark as B2 or B0
### Part (b)(i):
$\dfrac{(x-4)^2}{4} + (mx)^2 = 1$ | M1 | Substitute $y=mx$ to eliminate $y$
$(x-4)^2 + 4(mx)^2 = 4 \Rightarrow x^2 - 8x + 16 + 4(mx)^2 = 4$ | A1 | Eliminate fractions correctly and expand $(x-4)^2$ correctly
$\Rightarrow x^2 - 8x + 16 + 4m^2x^2 - 4 = 0$
$\Rightarrow (1+4m^2)x^2 - 8x + 12 = 0$ | A1 | CSO AG
### Part (b)(ii):
Discriminant $b^2 - 4ac$: $\{(-8)^2 - 4(1+4m^2)(12)\}$ | M1 | $b^2-4ac$ in terms of $m$, condone one sign or copying error OE
For tangency, $(-8)^2 - 4(1+4m^2)(12) = 0$ | A1 | A correct equation with $m^2$ being the only unknown at any stage.
$192m^2 - 16\ (=0)$ | A1 | OE eg $12m^2-1(=0)$ OE PI by a correct value for $m$ condoning wrong sign
$(m>0\ \text{so})\ m = \dfrac{1}{\sqrt{12}}$ | A1 | ACF of an exact value for $m$. eg $\frac{1}{2\sqrt{3}}, \frac{\sqrt{3}}{6}$. Dep on prev 3 marks
### Part (b)(iii):
$\left(1 + 4\times\left\{\dfrac{1}{\sqrt{12}}\right\}^2\right)x^2 - 8x + 12\ (=0)$ | M1 | Subst value for $m$ in LHS of eqn (b)(i); ft on c's value of $m$
$\dfrac{4}{3}x^2 - 8x + 12 = 0;\ \ 4x^2 - 24x + 36 = 0$
$x^2 - 6x + 9 = 0$ | m1 | Valid method to solve a correct quadratic equation; as far as either correct subst into quadratic formula with $b^2-4ac$ evaluated to 0, or correct factorisation or correct value of $x$ after $\frac{4}{3}x^2-8x+12=0$, or better seen; OE, correct use of $-b/2a$
$x = \dfrac{-(-8)\pm\sqrt{0}}{\frac{8}{3}};\ (x-3)^2\ (=0)$
$x = 3$ | A1 | Must see earlier justification
Coordinates of $P$ are $\left(3,\ \dfrac{3}{\sqrt{12}}\right)$ | A1 | Correct coordinates with the $y$-coord in any correct exact form eg $\frac{\sqrt{3}}{2}$. NMS SC1 for $\left(3, \frac{3}{\sqrt{12}}\right)$9 An ellipse is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{cf9337b9-b766-4ce5-967c-5d7522e2aa42-5_453_633_365_699}
The ellipse intersects the $x$-axis at the points $A$ and $B$. The equation of the ellipse is
$$\frac { ( x - 4 ) ^ { 2 } } { 4 } + y ^ { 2 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinates of $A$ and $B$.
\item The line $y = m x ( m > 0 )$ is a tangent to the ellipse, with point of contact $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of $P$ satisfies the equation
$$\left( 1 + 4 m ^ { 2 } \right) x ^ { 2 } - 8 x + 12 = 0$$
\item Hence find the exact value of $m$.
\item Find the coordinates of $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2013 Q9 [13]}}