| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Standard +0.3 This is a standard FP1 conic sections question requiring substitution of a line into a parabola equation, applying the tangency condition (discriminant = 0), and finding contact points. While it involves multiple steps, each follows a routine procedure with clear guidance, making it slightly easier than average for Further Maths content. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Parabola with vertex at \((2,0)\), opening rightward, correct shape | B1, B1 | B1 shape, B1 vertex/position |
| Answer | Marks |
|---|---|
| Two tangents from \((-2,0)\) drawn touching parabola symmetrically about x-axis | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = m(x+2)\); substitute into \(y^2 = x-2\): \(m^2(x+2)^2 = x-2\); \(m^2(x^2+4x+4)=x-2\); \(m^2x^2+4m^2x+4m^2-x+2=0\); \(m^2x^2+(4m^2-1)x+(4m^2+2)=0\) | M1, M1, A1 | M1 substitution, M1 expansion, A1 correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Equal roots: discriminant \(= 0\); \((4m^2-1)^2 - 4\cdot m^2\cdot(4m^2+2)=0\); \(16m^4-8m^2+1-16m^4-8m^2=0\); \(1-16m^2=0\); \(16m^2=1\) | M1, M1, A1 | M1 discriminant condition, M1 expansion, A1 shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = \pm\frac{1}{4}\); \(x = \frac{-(4m^2-1)}{2m^2} = \frac{0}{2m^2}\)... using equal roots: \(x = \frac{-(4m^2-1)}{2m^2}\); with \(16m^2=1\), \(m^2=\frac{1}{16}\): \(x=\frac{-(4/16-1)}{2/16}=\frac{3/4}{1/8}=6\); \(y=m(x+2)=\pm\frac{1}{4}(8)=\pm 2\); Points: \((6,2)\) and \((6,-2)\) | M1, A1, A1 | M1 finding x, A1 one point, A1 both points |
# Question 9:
**(a)(i)**
| Parabola with vertex at $(2,0)$, opening rightward, correct shape | B1, B1 | B1 shape, B1 vertex/position |
**(a)(ii)**
| Two tangents from $(-2,0)$ drawn touching parabola symmetrically about x-axis | B1, B1 | |
**(b)(i)**
| $y = m(x+2)$; substitute into $y^2 = x-2$: $m^2(x+2)^2 = x-2$; $m^2(x^2+4x+4)=x-2$; $m^2x^2+4m^2x+4m^2-x+2=0$; $m^2x^2+(4m^2-1)x+(4m^2+2)=0$ | M1, M1, A1 | M1 substitution, M1 expansion, A1 correct equation |
**(b)(ii)**
| Equal roots: discriminant $= 0$; $(4m^2-1)^2 - 4\cdot m^2\cdot(4m^2+2)=0$; $16m^4-8m^2+1-16m^4-8m^2=0$; $1-16m^2=0$; $16m^2=1$ | M1, M1, A1 | M1 discriminant condition, M1 expansion, A1 shown |
**(b)(iii)**
| $m = \pm\frac{1}{4}$; $x = \frac{-(4m^2-1)}{2m^2} = \frac{0}{2m^2}$... using equal roots: $x = \frac{-(4m^2-1)}{2m^2}$; with $16m^2=1$, $m^2=\frac{1}{16}$: $x=\frac{-(4/16-1)}{2/16}=\frac{3/4}{1/8}=6$; $y=m(x+2)=\pm\frac{1}{4}(8)=\pm 2$; Points: $(6,2)$ and $(6,-2)$ | M1, A1, A1 | M1 finding x, A1 one point, A1 both points |9 A parabola $P$ has equation $y ^ { 2 } = x - 2$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the parabola $P$.
\item On your sketch, draw the two tangents to $P$ which pass through the point $( - 2,0 )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that, if the line $y = m ( x + 2 )$ intersects $P$, then the $x$-coordinates of the points of intersection must satisfy the equation
$$m ^ { 2 } x ^ { 2 } + \left( 4 m ^ { 2 } - 1 \right) x + \left( 4 m ^ { 2 } + 2 \right) = 0$$
\item Show that, if this equation has equal roots, then
$$16 m ^ { 2 } = 1$$
\item Hence find the coordinates of the points at which the tangents to $P$ from the point $( - 2,0 )$ touch the parabola $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2010 Q9 [13]}}