| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| 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 implicit differentiation of a parabola, finding tangent equations using parametric forms, solving simultaneous equations with perpendicularity conditions, and calculating triangle area. While the techniques are standard for FP1, the combination of parametric coordinates, perpendicular tangent conditions, and multi-step algebraic manipulation makes this moderately challenging—harder than typical A-level Pure questions but routine for Further Maths students. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y^2 = 4x \Rightarrow 2y\frac{dy}{dx} = 4\) | M1A1 | \(x^{\frac{1}{2}} \to x^{-\frac{1}{2}}\); or \(ky\frac{dy}{dx}=c\); or \(\frac{dy}{dt}\times\frac{1}{\frac{dx}{dt}}\) |
| At \(P\): \(\frac{dy}{dx} = \frac{1}{p}\) | A1 | Accurate differentiation |
| \(y - 2p = \frac{1}{p}(x-p^2)\) | M1A1 | \(y-2p =\) their \(m(x-p^2)\); \(m\) must be function of \(p\) from calculus |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \((-1,2)\): \(2-2p = \frac{1}{p}(-1-p^2)\) | M1 | Substitute coordinates of \(R\) into tangent |
| \(p^2 - 2p - 1 = 0\) | A1 | |
| \(p = 1\pm\sqrt{2}\) | M1 | Solving 3-term quadratic |
| \(p = 1+\sqrt{2},\quad q = 1-\sqrt{2}\) | A1 | \(1\pm\sqrt{2}\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(PR^2 = 32+16\sqrt{2},\quad QR^2 = 32-16\sqrt{2}\) | M1A1 | Attempt to find distance between \(P\) and \(R\) or \(Q\) and \(R\) using formula or Pythagoras |
| Area of \(PQR = \frac{1}{2}PR\cdot QR = 8\sqrt{2}\) | M1A1 | Using \(\frac{1}{2}bh\) on \(PQR\); accept awrt 11.3 |
# Question 6:
## Part 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^2 = 4x \Rightarrow 2y\frac{dy}{dx} = 4$ | M1A1 | $x^{\frac{1}{2}} \to x^{-\frac{1}{2}}$; or $ky\frac{dy}{dx}=c$; or $\frac{dy}{dt}\times\frac{1}{\frac{dx}{dt}}$ |
| At $P$: $\frac{dy}{dx} = \frac{1}{p}$ | A1 | Accurate differentiation |
| $y - 2p = \frac{1}{p}(x-p^2)$ | M1A1 | $y-2p =$ their $m(x-p^2)$; $m$ must be function of $p$ from calculus |
## Part 6(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $(-1,2)$: $2-2p = \frac{1}{p}(-1-p^2)$ | M1 | Substitute coordinates of $R$ into tangent |
| $p^2 - 2p - 1 = 0$ | A1 | |
| $p = 1\pm\sqrt{2}$ | M1 | Solving 3-term quadratic |
| $p = 1+\sqrt{2},\quad q = 1-\sqrt{2}$ | A1 | $1\pm\sqrt{2}$ seen |
## Part 6(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PR^2 = 32+16\sqrt{2},\quad QR^2 = 32-16\sqrt{2}$ | M1A1 | Attempt to find distance between $P$ and $R$ or $Q$ and $R$ using formula or Pythagoras |
| Area of $PQR = \frac{1}{2}PR\cdot QR = 8\sqrt{2}$ | M1A1 | Using $\frac{1}{2}bh$ on $PQR$; accept awrt 11.3 |
---6. A curve $C$ is in the form of a parabola with equation $y ^ { 2 } = 4 x$.\\
$P \left( p ^ { 2 } , 2 p \right)$ and $Q \left( q ^ { 2 } , 2 q \right)$ are points on $C$ where $p > q$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the tangent to $C$ at $P$.\\
(5)
\item The tangent at $P$ and the tangent at $Q$ are perpendicular and intersect at the point $R ( - 1,2 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the exact value of $p$ and the exact value of $q$.
\item Find the area of the triangle $P Q R$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q6 [13]}}