| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola tangent equation derivation |
| Difficulty | Moderate -0.3 This is a standard FP1 parabola question requiring routine application of formulas. Part (a) is direct recall (directrix of y²=4ax), part (b) uses the standard parametric differentiation method taught in all FP1 courses, and part (c) involves substituting a specific point and solving simultaneous equations. While it requires multiple techniques, each step follows a well-practiced procedure with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y^2 = 4ax \Rightarrow a = \frac{48}{4} = 12\) | M1 | Using \(y^2 = 4ax\) to find \(a\) |
| Directrix: \(x + 12 = 0\) | A1 oe | \(x + 12 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2}\sqrt{48}\,x^{-\frac{1}{2}}\) or implicitly \(2y\frac{dy}{dx}=48\) or chain rule \(\frac{dy}{dx}=24\times\frac{1}{24t}\) | M1 | \(\frac{dy}{dx} = \pm kx^{-\frac{1}{2}}\), or \(ky\frac{dy}{dx}=c\), or \(\frac{dy}{dt}\times\frac{1}{\frac{dx}{dt}}\) |
| When \(x=12t^2\): \(\frac{dy}{dx} = \frac{1}{t}\) | A1 | \(\frac{dy}{dx} = \frac{1}{t}\) |
| T: \(y - 24t = \frac{1}{t}(x - 12t^2)\) | M1 | Applies \(y - 24t = m_T(x-12t^2)\) or \(y = m_T x + c\) using \(x=12t^2\), \(y=24t\). \(m_T\) must be a function of \(t\) |
| T: \(x - ty + 12t^2 = 0\) | A1 cso* | Correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Compare \(P(12t^2, 24t)\) with \((3,12)\): \(t = \frac{1}{2}\) | B1 | \(t = \frac{1}{2}\) |
| \(t=\frac{1}{2}\) into T: \(x - \frac{1}{2}y + 3 = 0\) | M1 | Substitutes their \(t\) into T |
| At \(X\), \(x = -12\): \(-12 - \frac{1}{2}y + 3 = 0\) | M1 | Substitutes their \(x\) from (a) into T |
| Coordinates of \(X\) are \((-12, -18)\) | A1 | \((-12, -18)\) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^2 = 4ax \Rightarrow a = \frac{48}{4} = 12$ | M1 | Using $y^2 = 4ax$ to find $a$ |
| Directrix: $x + 12 = 0$ | A1 oe | $x + 12 = 0$ |
**(2 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2}\sqrt{48}\,x^{-\frac{1}{2}}$ or implicitly $2y\frac{dy}{dx}=48$ or chain rule $\frac{dy}{dx}=24\times\frac{1}{24t}$ | M1 | $\frac{dy}{dx} = \pm kx^{-\frac{1}{2}}$, or $ky\frac{dy}{dx}=c$, or $\frac{dy}{dt}\times\frac{1}{\frac{dx}{dt}}$ |
| When $x=12t^2$: $\frac{dy}{dx} = \frac{1}{t}$ | A1 | $\frac{dy}{dx} = \frac{1}{t}$ |
| **T**: $y - 24t = \frac{1}{t}(x - 12t^2)$ | M1 | Applies $y - 24t = m_T(x-12t^2)$ or $y = m_T x + c$ using $x=12t^2$, $y=24t$. $m_T$ must be a function of $t$ |
| **T**: $x - ty + 12t^2 = 0$ | A1 cso* | Correct solution |
**(4 marks)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Compare $P(12t^2, 24t)$ with $(3,12)$: $t = \frac{1}{2}$ | B1 | $t = \frac{1}{2}$ |
| $t=\frac{1}{2}$ into **T**: $x - \frac{1}{2}y + 3 = 0$ | M1 | Substitutes their $t$ into **T** |
| At $X$, $x = -12$: $-12 - \frac{1}{2}y + 3 = 0$ | M1 | Substitutes their $x$ from (a) into **T** |
| Coordinates of $X$ are $(-12, -18)$ | A1 | $(-12, -18)$ |
**(4 marks)**8. The parabola $C$ has equation $y ^ { 2 } = 48 x$.
The point $P \left( 12 t ^ { 2 } , 24 t \right)$ is a general point on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the directrix of $C$.
\item Show that the equation of the tangent to $C$ at $P \left( 12 t ^ { 2 } , 24 t \right)$ is
$$x - t y + 12 t ^ { 2 } = 0$$
The tangent to $C$ at the point $( 3,12 )$ meets the directrix of $C$ at the point $X$.
\item Find the coordinates of $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q8 [10]}}