| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Standard +0.8 This is a Further Maths FP1 parabola locus problem requiring knowledge of focus-directrix properties, parametric coordinates, and algebraic manipulation to prove a locus equation. Parts (a)-(b) are routine recall, but part (c) requires systematic elimination of parameters with the constraint pq=-1, which is non-trivial but follows standard FP1 techniques. Slightly above average difficulty for Further Maths content. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((a,0)\) | B1 | Correct coordinates of focus |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(SP=ap^2+a\); e.g. \(SP=\sqrt{4a^2p^2+(ap^2-a)^2}=\ldots=ap^2+a\) | B1 | Correct expression (focus-directrix property or Pythagoras) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(M\) has coordinates \(\left(\frac{ap^2+aq^2}{2},\frac{2ap+2aq}{2}\right)\) | B1 | Correct coordinates for midpoint |
| \(y^2=a^2(p^2+2pq+q^2)\) | M1 | Squares \(y\)-coordinate of midpoint |
| \(y^2=a^2(p^2-2+q^2)\) | A1 | Uses \(pq=-1\) to obtain correct expression for \(y^2\) |
| \(2a(x-a)=2a\left(\frac{1}{2}ap^2+\frac{1}{2}aq^2-a\right)=a^2(p^2+q^2-2)\) | M1 | Attempts \(2a(x-a)\) using \(x\)-coordinate of midpoint |
| \(\Rightarrow y^2=2a(x-a)\) | A1* | Fully correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(M\) has coordinates \(\left(\frac{ap^2+aq^2}{2},\frac{2ap+2aq}{2}\right)\) | B1 | Correct midpoint coordinates |
| \(\frac{y}{a}=p+q\) | M1 | Uses \(y\)-coordinate to find \(p+q\) |
| \(\frac{y^2}{a^2}=p^2+q^2+2pq=p^2+q^2-2\) | A1 | Squares and uses \(pq=-1\) |
| \(\frac{2x}{a}=p^2+q^2\) | M1 | Uses \(x\)-coordinate to find \(p^2+q^2\) |
| \(\frac{y^2}{a^2}=\frac{2x}{a}-2\Rightarrow y^2=2a(x-a)\) | A1* | Fully correct completion |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(a,0)$ | B1 | Correct coordinates of focus |
---
## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $SP=ap^2+a$; e.g. $SP=\sqrt{4a^2p^2+(ap^2-a)^2}=\ldots=ap^2+a$ | B1 | Correct expression (focus-directrix property or Pythagoras) |
---
## Question 4(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $M$ has coordinates $\left(\frac{ap^2+aq^2}{2},\frac{2ap+2aq}{2}\right)$ | B1 | Correct coordinates for midpoint |
| $y^2=a^2(p^2+2pq+q^2)$ | M1 | Squares $y$-coordinate of midpoint |
| $y^2=a^2(p^2-2+q^2)$ | A1 | Uses $pq=-1$ to obtain correct expression for $y^2$ |
| $2a(x-a)=2a\left(\frac{1}{2}ap^2+\frac{1}{2}aq^2-a\right)=a^2(p^2+q^2-2)$ | M1 | Attempts $2a(x-a)$ using $x$-coordinate of midpoint |
| $\Rightarrow y^2=2a(x-a)$ | A1* | Fully correct completion |
**Alternative for (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $M$ has coordinates $\left(\frac{ap^2+aq^2}{2},\frac{2ap+2aq}{2}\right)$ | B1 | Correct midpoint coordinates |
| $\frac{y}{a}=p+q$ | M1 | Uses $y$-coordinate to find $p+q$ |
| $\frac{y^2}{a^2}=p^2+q^2+2pq=p^2+q^2-2$ | A1 | Squares and uses $pq=-1$ |
| $\frac{2x}{a}=p^2+q^2$ | M1 | Uses $x$-coordinate to find $p^2+q^2$ |
| $\frac{y^2}{a^2}=\frac{2x}{a}-2\Rightarrow y^2=2a(x-a)$ | A1* | Fully correct completion |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1e5324f5-a9bc-4041-bfbb-cb940417ea63-12_611_608_274_715}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the parabola $C$ with equation $y ^ { 2 } = 4 a x$, where $a$ is a positive constant. The point $S$ is the focus of $C$ and the point $P \left( a p ^ { 2 } \right.$, 2ap) lies on $C$ where $p > 0$
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $S$.
\item Write down the length of SP in terms of $a$ and $p$.
The point $Q \left( a q ^ { 2 } , 2 a q \right)$, where $p \neq q$, also lies on $C$.\\
The point $M$ is the midpoint of $P Q$.\\
Given that $p q = - 1$
\item prove that, as $P$ varies, the locus of $M$ has equation
$$y ^ { 2 } = 2 a ( x - a )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 AS 2020 Q4 [7]}}