| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Chord midpoint locus |
| Difficulty | Standard +0.8 This is a multi-part Further Maths conic sections question requiring algebraic manipulation to find intersection points, use of chord midpoint formulas, and elimination of a parameter to find a locus. While systematic, it demands careful algebra across multiple steps and understanding of parametric elimination—moderately challenging for Further Maths students. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{x^2}{9} + \frac{(kx-3)^2}{4} = 1 \Rightarrow 4x^2 + 9(k^2x^2 - 6kx + 9) = 36\) | M1 | Substitutes to obtain a quadratic in \(x\) and eliminates fractions |
| \((9k^2+4)x^2 - 54kx + 45 = 0\) * | A1* | Correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = \frac{1}{2}\left(\frac{54k}{9k^2+4}\right) = \frac{27k}{9k^2+4}\) OR \(x = \frac{54k \pm \sqrt{\text{discriminant}}}{2(9k^2+4)}\) | M1 | Uses \(\frac{1}{2}\) sum of roots for \(x\)-coordinate OR solve by formula, add 2 roots and halve; must reach \(x_m\); allow errors in discriminant |
| \(y = k\left(\frac{27k}{9k^2+4}\right) - 3 = \frac{27k^2 - 27k^2 - 12}{9k^2+4} = -\frac{12}{9k^2+4}\) | dM1 | Uses straight line equation to find \(y\) as single fraction; depends on first M mark of (b) |
| \(x = \frac{27k}{9k^2+4}\), \(y = -\frac{12}{9k^2+4}\) | A1 | Fully correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^2 = \frac{729k^2}{(9k^2+4)^2} \Rightarrow x^2 + py^2 = \frac{729k^2 + 144p}{(9k^2+4)^2}\) | M1 | Obtains expression for \(x^2 + py^2\) using coordinates from (b) and obtains common denominator |
| \(\frac{729k^2+144p}{(9k^2+4)^2} = -\frac{12q}{(9k^2+4)} \Rightarrow 729k^2 + 144p = -12q(9k^2+4)\) then \(729k^2 + 144p = 81\left(9k^2 + \frac{16}{9}p\right) \Rightarrow \frac{16}{9}p = 4 \Rightarrow p = \ldots\) | dM1 | Correct strategy to obtain value for \(p\) or \(q\); depends on first M mark of (c) |
| \(p = \frac{9}{4}\) or \(q = -\frac{27}{4}\) | A1 | Correct value (or for \(q\) if found first) |
| \(-12q = 81 \Rightarrow q = \ldots\) | ddM1 | Correct strategy for second variable; depends on both previous M marks |
| \(x^2 + \frac{9}{4}y^2 = -\frac{27}{4}y\); \(p = \frac{9}{4}\) and \(q = -\frac{27}{4}\) | A1 | Both values correct (can be embedded in equation) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = \frac{27k}{9k^2+4}\), \(y = -\frac{12}{9k^2+4} \Rightarrow \frac{x}{y} = -\frac{27k}{12} \Rightarrow k = -\frac{4x}{9y}\) | M1 | Obtains \(k\) in terms of \(x\) and \(y\) using coordinates from (b) |
| \(k = -\frac{4x}{9y} \Rightarrow y = -\frac{12}{9\left(\frac{16x^2}{81y^2}\right)+4}\) or \(x = \frac{27\left(-\frac{4x}{9y}\right)}{9\left(\frac{16x^2}{81y^2}\right)+4}\) | dM1A1 | dM1: Substitutes \(k\) into \(y\) or \(x\) to obtain Cartesian equation; A1: Any correct Cartesian equation |
| \(\Rightarrow x^2 + \frac{9}{4}y^2 = -\frac{27}{4}y\) | ddM1, A1 | ddM1: Rearranges to required form; depends on both previous M marks. A1: Correct equation or correct values stated |
# Question 9:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{x^2}{9} + \frac{(kx-3)^2}{4} = 1 \Rightarrow 4x^2 + 9(k^2x^2 - 6kx + 9) = 36$ | M1 | Substitutes to obtain a quadratic in $x$ and eliminates fractions |
| $(9k^2+4)x^2 - 54kx + 45 = 0$ * | A1* | Correct proof with no errors |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = \frac{1}{2}\left(\frac{54k}{9k^2+4}\right) = \frac{27k}{9k^2+4}$ OR $x = \frac{54k \pm \sqrt{\text{discriminant}}}{2(9k^2+4)}$ | M1 | Uses $\frac{1}{2}$ sum of roots for $x$-coordinate OR solve by formula, add 2 roots and halve; must reach $x_m$; allow errors in discriminant |
| $y = k\left(\frac{27k}{9k^2+4}\right) - 3 = \frac{27k^2 - 27k^2 - 12}{9k^2+4} = -\frac{12}{9k^2+4}$ | dM1 | Uses straight line equation to find $y$ as single fraction; depends on first M mark of (b) |
| $x = \frac{27k}{9k^2+4}$, $y = -\frac{12}{9k^2+4}$ | A1 | Fully correct work |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^2 = \frac{729k^2}{(9k^2+4)^2} \Rightarrow x^2 + py^2 = \frac{729k^2 + 144p}{(9k^2+4)^2}$ | M1 | Obtains expression for $x^2 + py^2$ using coordinates from (b) and obtains common denominator |
| $\frac{729k^2+144p}{(9k^2+4)^2} = -\frac{12q}{(9k^2+4)} \Rightarrow 729k^2 + 144p = -12q(9k^2+4)$ then $729k^2 + 144p = 81\left(9k^2 + \frac{16}{9}p\right) \Rightarrow \frac{16}{9}p = 4 \Rightarrow p = \ldots$ | dM1 | Correct strategy to obtain value for $p$ or $q$; depends on first M mark of (c) |
| $p = \frac{9}{4}$ or $q = -\frac{27}{4}$ | A1 | Correct value (or for $q$ if found first) |
| $-12q = 81 \Rightarrow q = \ldots$ | ddM1 | Correct strategy for second variable; depends on both previous M marks |
| $x^2 + \frac{9}{4}y^2 = -\frac{27}{4}y$; $p = \frac{9}{4}$ and $q = -\frac{27}{4}$ | A1 | Both values correct (can be embedded in equation) |
**Way 2 (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = \frac{27k}{9k^2+4}$, $y = -\frac{12}{9k^2+4} \Rightarrow \frac{x}{y} = -\frac{27k}{12} \Rightarrow k = -\frac{4x}{9y}$ | M1 | Obtains $k$ in terms of $x$ and $y$ using coordinates from (b) |
| $k = -\frac{4x}{9y} \Rightarrow y = -\frac{12}{9\left(\frac{16x^2}{81y^2}\right)+4}$ or $x = \frac{27\left(-\frac{4x}{9y}\right)}{9\left(\frac{16x^2}{81y^2}\right)+4}$ | dM1A1 | dM1: Substitutes $k$ into $y$ or $x$ to obtain Cartesian equation; A1: Any correct Cartesian equation |
| $\Rightarrow x^2 + \frac{9}{4}y^2 = -\frac{27}{4}y$ | ddM1, A1 | ddM1: Rearranges to required form; depends on both previous M marks. A1: Correct equation or correct values stated |\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1$$
The line $l$ has equation $y = k x - 3$, where $k$ is a constant.\\
Given that $E$ and $l$ meet at 2 distinct points $P$ and $Q$\\
(a) show that the $x$ coordinates of $P$ and $Q$ are solutions of the equation
$$\left( 9 k ^ { 2 } + 4 \right) x ^ { 2 } - 54 k x + 45 = 0$$
The point $M$ is the midpoint of $P Q$\\
(b) Determine, in simplest form in terms of $k$, the coordinates of $M$\\
(c) Hence show that, as $k$ varies, $M$ lies on the curve with equation
$$x ^ { 2 } + p y ^ { 2 } = q y$$
where $p$ and $q$ are constants to be determined.
\hfill \mbox{\textit{Edexcel F3 2022 Q9 [10]}}