| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring parametric coordinate manipulation, finding normals to a parabola, solving simultaneous equations to find intersection points, and deriving a locus equation by eliminating parameters. While the algebraic steps are guided by the 'show that' structure, part (c) requires significant algebraic manipulation and insight to eliminate two parameters (p and q) using the constraint that l passes through (12,0). This is more demanding than typical FP1 questions but follows standard locus techniques. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07m Tangents and normals: gradient and equations4.03d Linear transformations 2D: reflection, rotation, enlargement, shear |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Gradient of \(PQ = \frac{18q-18p}{9q^2-9p^2} = \frac{2}{p+q}\) | B1 | Deduces gradient is \(\frac{2}{p+q}\); may be implied by correct simplification of equation |
| Equation of \(l\): \(y - 18p = \frac{2}{p+q}(x - 9p^2)\) | M1 | Correct method for equation of line; gradient need not be simplified/correct |
| Leading to \((p+q)y = 2(x+9pq)\)* | A1\* | Completes to correct equation with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Complete method to find equation of both normals and attempt to solve simultaneously | M1 | Must find both normals and attempt simultaneous solution |
| \(2y\frac{dy}{dx} = 36 \Rightarrow m_T = \frac{36}{36p} \Rightarrow m_N = -p\) | B1 | Correct gradient of normal from any correct method |
| Normal at \(P\): \(y-18p = -p(x-9p^2)\) or normal at \(Q\): \(y-18q = -q(x-9q^2)\) | M1 | Full correct method for equation of at least one normal with justification of gradient |
| Both normals correct: \(y = -px + 9p^3 + 18p\) and \(y = -qx + 9q^3 + 18q\) | A1 | Deduces equation of second normal — both correct |
| \(18p - px + 9p^3 - 18q = -qx + 9q^3 \Rightarrow x = ...\) | M1 | Solves two normal equations simultaneously leading to \(x=...\) or \(y=...\) |
| Need to show \((9p^3-9q^3+18p-18q) = (9p^2+q^2+pq+2)(p-q)\), leading to \(x_A = 9(p^2+q^2+pq+2)\)* | A1\* | Must show factorisation leading to correct \(x\)-coordinate with no errors; could use long division or factorising |
| \(y_A = -9pq(p+q)\)* | A1\* | Correct \(y\)-coordinate with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((12,0)\) on \(l \Rightarrow pq = -\frac{4}{3}\) | B1 | Uses condition on \(l\) to establish relationship between \(p\) and \(q\) |
| Hence \(x_A = 9\!\left(p^2+q^2+\frac{2}{3}\right)\) and \(y_A = 12(p+q)\) | M1 | Uses their relationship between \(p\) and \(q\) to simplify expressions |
| \(y^2 = 144(p^2+q^2+2pq) = 144\!\left(\frac{x}{9}-\frac{2}{3}+2\!\left(-\frac{4}{3}\right)\right)\) | M1 | Any complete method for relating \(x\) and \(y\) independently of \(p\) and \(q\) |
| \(y^2 = 16(x-30)\) or \(y^2 = 16x - 480\) | A1 | Correct final answer |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Gradient of $PQ = \frac{18q-18p}{9q^2-9p^2} = \frac{2}{p+q}$ | **B1** | Deduces gradient is $\frac{2}{p+q}$; may be implied by correct simplification of equation |
| Equation of $l$: $y - 18p = \frac{2}{p+q}(x - 9p^2)$ | **M1** | Correct method for equation of line; gradient need not be simplified/correct |
| Leading to $(p+q)y = 2(x+9pq)$* | **A1\*** | Completes to correct equation with no errors seen |
**(3 marks)**
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Complete method to find equation of both normals and attempt to solve simultaneously | **M1** | Must find both normals and attempt simultaneous solution |
| $2y\frac{dy}{dx} = 36 \Rightarrow m_T = \frac{36}{36p} \Rightarrow m_N = -p$ | **B1** | Correct gradient of normal from any correct method |
| Normal at $P$: $y-18p = -p(x-9p^2)$ or normal at $Q$: $y-18q = -q(x-9q^2)$ | **M1** | Full correct method for equation of at least one normal with justification of gradient |
| Both normals correct: $y = -px + 9p^3 + 18p$ and $y = -qx + 9q^3 + 18q$ | **A1** | Deduces equation of second normal — both correct |
| $18p - px + 9p^3 - 18q = -qx + 9q^3 \Rightarrow x = ...$ | **M1** | Solves two normal equations simultaneously leading to $x=...$ or $y=...$ |
| Need to show $(9p^3-9q^3+18p-18q) = (9p^2+q^2+pq+2)(p-q)$, leading to $x_A = 9(p^2+q^2+pq+2)$* | **A1\*** | Must show factorisation leading to correct $x$-coordinate with no errors; could use long division or factorising |
| $y_A = -9pq(p+q)$* | **A1\*** | Correct $y$-coordinate with no errors seen |
**(7 marks)**
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(12,0)$ on $l \Rightarrow pq = -\frac{4}{3}$ | **B1** | Uses condition on $l$ to establish relationship between $p$ and $q$ |
| Hence $x_A = 9\!\left(p^2+q^2+\frac{2}{3}\right)$ and $y_A = 12(p+q)$ | **M1** | Uses their relationship between $p$ and $q$ to simplify expressions |
| $y^2 = 144(p^2+q^2+2pq) = 144\!\left(\frac{x}{9}-\frac{2}{3}+2\!\left(-\frac{4}{3}\right)\right)$ | **M1** | Any complete method for relating $x$ and $y$ independently of $p$ and $q$ |
| $y^2 = 16(x-30)$ or $y^2 = 16x - 480$ | **A1** | Correct final answer |
**(4 marks)**
**(14 marks total)**\begin{enumerate}
\item The points $P \left( 9 p ^ { 2 } , 18 p \right)$ and $Q \left( 9 q ^ { 2 } , 18 q \right) , p \neq q$, lie on the parabola $C$ with equation
\end{enumerate}
$$y ^ { 2 } = 36 x$$
The line $l$ passes through the points $P$ and $Q$\\
(a) Show that an equation for the line $l$ is
$$( p + q ) y = 2 ( x + 9 p q )$$
The normal to $C$ at $P$ and the normal to $C$ at $Q$ meet at the point $A$.\\
(b) Show that the coordinates of $A$ are
$$\left( 9 \left( p ^ { 2 } + q ^ { 2 } + p q + 2 \right) , - 9 p q ( p + q ) \right)$$
Given that the points $P$ and $Q$ vary such that $l$ always passes through the point $( 12,0 )$\\
(c) find, in the form $y ^ { 2 } = \mathrm { f } ( x )$, an equation for the locus of $A$, giving $\mathrm { f } ( x )$ in simplest form.
\hfill \mbox{\textit{Edexcel FP1 2020 Q7 [14]}}