| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Standard +0.3 This is a straightforward Further Pure question on parabolas covering standard transformations (translation, reflection) and tangency conditions using the discriminant. Part (c)(i) is routine substitution and algebra, while finding the tangent condition (discriminant = 0) is a textbook application. The question requires multiple techniques but no novel insight, making it slightly easier than average for FP1 material, which itself is more challenging than standard A-level. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Good attempt at sketch | M1 | |
| Correct at origin | A1 | |
| (b)(i) \(y\) replaced by \(y - 2\) | B1 | |
| Equation is \((y - 2)^2 = 12x\) | B1√ | ft \(y + 2\) for \(y - 2\) |
| (ii) Equation is \(x^2 = 12y\) | B1 | |
| (c)(i) \((x + c)^2 = x^2 + 2cx + c^2 ... = 12x\) | M1 | |
| Hence result | A1 | Convincingly shown (AG) |
| (ii) Tangent if \((2c - 12)^2 - 4c^2 = 0\) | M1 | |
| ie if \(-48c + 144 = 0\) so \(c = 3\) | A1 | |
| (iii) \(x^2 - 6x + 9 = 0\) | M1 | |
| \(x = 3\), \(y = 6\) | A1 | |
| (iv) \(c = 4 \Rightarrow\) discriminant \(= -48 < 0\) | M1A1 | OE |
| So line does not intersect curve | A1 |
**(a)** Good attempt at sketch | M1
Correct at origin | A1 |
**(b)(i)** $y$ replaced by $y - 2$ | B1
Equation is $(y - 2)^2 = 12x$ | B1√ | ft $y + 2$ for $y - 2$
**(ii)** Equation is $x^2 = 12y$ | B1 |
**(c)(i)** $(x + c)^2 = x^2 + 2cx + c^2 ... = 12x$ | M1
Hence result | A1 | Convincingly shown (AG)
**(ii)** Tangent if $(2c - 12)^2 - 4c^2 = 0$ | M1
ie if $-48c + 144 = 0$ so $c = 3$ | A1 |
**(iii)** $x^2 - 6x + 9 = 0$ | M1
$x = 3$, $y = 6$ | A1 |
**(iv)** $c = 4 \Rightarrow$ discriminant $= -48 < 0$ | M1A1 | OE
So line does not intersect curve | A1 |
**Total for Q8: 15 marks**
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## TOTAL: 75 marks8 A curve has equation $y ^ { 2 } = 12 x$.
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve.
\item \begin{enumerate}[label=(\roman*)]
\item The curve is translated by 2 units in the positive $y$ direction. Write down the equation of the curve after this translation.
\item The original curve is reflected in the line $y = x$. Write down the equation of the curve after this reflection.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that if the straight line $y = x + c$, where $c$ is a constant, intersects the curve $y ^ { 2 } = 12 x$, then the $x$-coordinates of the points of intersection satisfy the equation
$$x ^ { 2 } + ( 2 c - 12 ) x + c ^ { 2 } = 0$$
\item Hence find the value of $c$ for which the straight line is a tangent to the curve.
\item Using this value of $c$, find the coordinates of the point where the line touches the curve.
\item In the case where $c = 4$, determine whether the line intersects the curve or not.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2006 Q8 [15]}}