| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola tangent equation derivation |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring parametric parabola knowledge, implicit differentiation, and coordinate geometry with the directrix. Part (a) is routine substitution, part (b) requires careful differentiation of the parametric form, and part (c) demands solving simultaneous equations involving two tangents meeting on the directrix—requiring sustained algebraic manipulation and conceptual understanding beyond standard A-level. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(a = 4\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow y' = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) | M1 | Attempt \(y'\) |
| \(y' = \frac{1}{t}\), substituting \(x = 4t^2\) | M1 | |
| Tangent: \(y - 8t = \frac{1}{t}(x - 4t^2)\) | M1 | |
| \(yt = x + 4t^2\) | A1cso | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = -4\) | B1ft | |
| \(15t = -4 + 4t^2\), substitute \((-4, 15)\) | M1 | |
| \(4t^2 - 15t - 4 = 0\) | ||
| \((4t+1)(t-4) = 0\) | M1 | Attempt to solve |
| \(t = 4\) or \(t = -\frac{1}{4}\) | A1 | |
| \(A = (64, 32)\), \(B = (\frac{1}{4}, -2)\) | M1 A1 A1 | (7 marks) |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $a = 4$ | B1 | **(1 mark)** |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow y' = a^{\frac{1}{2}}x^{-\frac{1}{2}}$ | M1 | Attempt $y'$ |
| $y' = \frac{1}{t}$, substituting $x = 4t^2$ | M1 | |
| Tangent: $y - 8t = \frac{1}{t}(x - 4t^2)$ | M1 | |
| $yt = x + 4t^2$ | A1cso | **(4 marks)** |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = -4$ | B1ft | |
| $15t = -4 + 4t^2$, substitute $(-4, 15)$ | M1 | |
| $4t^2 - 15t - 4 = 0$ | | |
| $(4t+1)(t-4) = 0$ | M1 | Attempt to solve |
| $t = 4$ or $t = -\frac{1}{4}$ | A1 | |
| $A = (64, 32)$, $B = (\frac{1}{4}, -2)$ | M1 A1 A1 | **(7 marks)** |
---7. The parabola $C$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a constant.
The point $\left( 4 t ^ { 2 } , 8 t \right)$ is a general point on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item Show that the equation for the tangent to $C$ at the point $\left( 4 t ^ { 2 } , 8 t \right)$ is
$$y t = x + 4 t ^ { 2 } .$$
The tangent to $C$ at the point $A$ meets the tangent to $C$ at the point $B$ on the directrix of $C$ when $y = 15$.
\item Find the coordinates of $A$ and the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q7 [12]}}