Question 3 - A-Level Maths

Edexcel FP3 2013 June — Question 3 8 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2013
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Ellipse locus problems
Difficulty	Challenging +1.2 This is a standard FP3 locus problem requiring parametric representation of an ellipse, coordinate geometry to find the midpoint, and algebraic manipulation to eliminate the parameter. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the approach is methodical and follows a well-established template for ellipse locus questions with no novel geometric insight required.
Spec	1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle

The point $P$ lies on the ellipse $E$ with equation

$$\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 9 } = 1$$ $N$ is the foot of the perpendicular from point $P$ to the line $x = 8$ $M$ is the midpoint of $P N$.

Sketch the graph of the ellipse $E$, showing also the line $x = 8$ and a possible position for the line $P N$.
Find an equation of the locus of $M$ as $P$ moves around the ellipse.
Show that this locus is a circle and state its centre and radius.

Show mark scheme Show mark scheme source

Question 3:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Sketch: closed curve symmetrical about both axes, vertical line to right of curve, horizontal line from point on ellipse to vertical line with P and N marked	B1	A closed curve approximately symmetrical about both axes. A vertical line to the right of the curve. A horizontal line from any point on the ellipse to the vertical line with both P and N clearly marked.
(1)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$M$ is $\left(\frac{x+8}{2}, y\right) = (X,Y)$ or $\left(\frac{6\cos\theta+8}{2}, 3\sin\theta\right) = (X,Y)$	M1	Finds the mid-point of PN
A1	Correct mid-point
$\frac{(2X-8)^2}{36} + \frac{Y^2}{9} = 1$	M1	Attempt Cartesian equation
A1	Correct equation
(4)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Circle because equation may be written $(x-4)^2 + y^2 = 3^2$	B1ft	Convincing argument — allow follow through provided they do have a circle. Can be implied by their centre and radius.
The centre is $(4, 0)$ and the radius is $3$	M1	Use their circle equation to find centre and radius
A1	Correct centre and radius
(3)
Total 8

Special Case: In (b) if locus assumed circle, intercepts on $x$-axis as $(1,0)$ and $(7,0)$ deducing centre $(4,0)$ and radius $3$: scores no marks in (b) but allow recovery in (c).

# Question 3:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch: closed curve symmetrical about both axes, vertical line to right of curve, horizontal line from point on ellipse to vertical line with P and N marked | B1 | A closed curve approximately symmetrical about both axes. A vertical line to the right of the curve. A horizontal line from any point on the ellipse to the vertical line **with both P and N clearly marked**. |
| | **(1)** | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $M$ is $\left(\frac{x+8}{2}, y\right) = (X,Y)$ or $\left(\frac{6\cos\theta+8}{2}, 3\sin\theta\right) = (X,Y)$ | M1 | Finds the mid-point of PN |
| | A1 | Correct mid-point |
| $\frac{(2X-8)^2}{36} + \frac{Y^2}{9} = 1$ | M1 | Attempt Cartesian equation |
| | A1 | Correct equation |
| | **(4)** | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle because equation may be written $(x-4)^2 + y^2 = 3^2$ | B1ft | Convincing argument — allow follow through provided they do have a circle. Can be implied by their centre and radius. |
| The centre is $(4, 0)$ and the radius is $3$ | M1 | Use their circle equation to find centre and radius |
| | A1 | Correct centre and radius |
| | **(3)** | |
| | **Total 8** | |

**Special Case:** In (b) if locus assumed circle, intercepts on $x$-axis as $(1,0)$ and $(7,0)$ deducing centre $(4,0)$ and radius $3$: scores no marks in (b) but allow recovery in (c).

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Show LaTeX source

\begin{enumerate}
  \item The point $P$ lies on the ellipse $E$ with equation
\end{enumerate}

$$\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 9 } = 1$$

$N$ is the foot of the perpendicular from point $P$ to the line $x = 8$\\
$M$ is the midpoint of $P N$.\\
(a) Sketch the graph of the ellipse $E$, showing also the line $x = 8$ and a possible position for the line $P N$.\\
(b) Find an equation of the locus of $M$ as $P$ moves around the ellipse.\\
(c) Show that this locus is a circle and state its centre and radius.\\

\hfill \mbox{\textit{Edexcel FP3 2013 Q3 [8]}}

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Q1 7 Q2 7 Q3 8 Q4 9 Q5 10 Q6 11 Q7 12 Q8 11

Edexcel FP3 2013 June — Question 3 8 marks

Question 3:

Part (a):

Part (b):

Part (c):

Special Case: In (b) if locus assumed circle, intercepts on \(x\)-axis as \((1,0)\) and \((7,0)\) deducing centre \((4,0)\) and radius \(3\): scores no marks in (b) but allow recovery in (c).

This paper (8 questions)