Question 3 - A-Level Maths

Edexcel FP1 AS 2023 June — Question 3 5 marks

Exam Board	Edexcel
Module	FP1 AS (Further Pure 1 AS)
Year	2023
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Chord midpoint locus
Difficulty	Challenging +1.2 This is a standard Further Pure 1 chord midpoint locus problem requiring substitution to find intersection points, solving a quadratic, finding the midpoint, and eliminating the parameter. While it involves multiple steps and algebraic manipulation, the techniques are routine for FP1 students with no novel geometric insight required. The structure is predictable and similar to textbook exercises, placing it moderately above average difficulty.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.02q Use intersection points: of graphs to solve equations 1.03g Parametric equations: of curves and conversion to cartesian

The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$ where $c$ is a positive constant.

The line $l$ has equation $x - 2 y = c$ The points $P$ and $Q$ are the points of intersection of $H$ and $l$

Determine, in terms of $c$, the coordinates of $P$ and the coordinates of $Q$ The point $R$ is the midpoint of $P Q$
Show that, as $C$ varies, the coordinates of $R$ satisfy the equation $$x y = - \frac { c ^ { 2 } } { a }$$ where $a$ is a constant to be determined.

Show mark scheme Show mark scheme source

Question 3:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$xy = c^2$ and $x - 2y = c \Rightarrow 2y^2 + cy = c^2$ or $\frac{1}{2}x^2 - cx = c^2$	M1	Solve simultaneously leading to quadratic in either $x$ or $y$ (and $c$)
$2y^2 + cy - c^2 = 0 \Rightarrow y = -c, \frac{c}{2} \Rightarrow x = ...$ or $x^2 - cx - 2c^2 = 0 \Rightarrow x = -c, 2c \Rightarrow y = ...$	dM1	Dependent on M1; solves quadratic to find at least one set of coordinates
$(-c,\ -c)$ and $\left(2c,\ \frac{c}{2}\right)$	A1	Deduces correct coordinates for both $P$ and $Q$; accept $x=...,\ y=...$ as long as paired

Alternative (parametric):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
General point $\left(ct, \frac{c}{t}\right) \Rightarrow ct - \frac{2c}{t} = c \Rightarrow ct^2 - 2c = ct$	M1	Uses parametric equations, substitutes into $x - 2y = c$, leading to quadratic in $t$
$\Rightarrow t^2 - t - 2 = 0 \Rightarrow t = 2, -1 \Rightarrow \left(2c, \frac{c}{2}\right)$ or $(-c, -c)$	dM1	Solves quadratic in $t$
$(-c,\ -c)$ and $\left(2c,\ \frac{c}{2}\right)$	A1	Correct coordinates for both points

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\text{Midpoint} = \left(\frac{-c+2c}{2},\ \frac{-c + \frac{c}{2}}{2}\right) = ...$	M1	Find midpoint of their $P$ and $Q$, provided not symmetric in $y$-axis
$x = \frac{c}{2}$ and $y = -\frac{c}{4} \Rightarrow xy = \frac{c}{2} \times -\frac{c}{4}$ leading to $xy = -\frac{c^2}{8}$	A1cso	Uses correct midpoint coordinates to show $xy = -\frac{c^2}{8}$; must be an equation, no contrary statements

## Question 3:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $xy = c^2$ and $x - 2y = c \Rightarrow 2y^2 + cy = c^2$ or $\frac{1}{2}x^2 - cx = c^2$ | M1 | Solve simultaneously leading to quadratic in either $x$ or $y$ (and $c$) |
| $2y^2 + cy - c^2 = 0 \Rightarrow y = -c, \frac{c}{2} \Rightarrow x = ...$ or $x^2 - cx - 2c^2 = 0 \Rightarrow x = -c, 2c \Rightarrow y = ...$ | dM1 | Dependent on M1; solves quadratic to find at least one set of coordinates |
| $(-c,\ -c)$ and $\left(2c,\ \frac{c}{2}\right)$ | A1 | Deduces correct coordinates for both $P$ and $Q$; accept $x=...,\ y=...$ as long as paired |

**Alternative (parametric):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| General point $\left(ct, \frac{c}{t}\right) \Rightarrow ct - \frac{2c}{t} = c \Rightarrow ct^2 - 2c = ct$ | M1 | Uses parametric equations, substitutes into $x - 2y = c$, leading to quadratic in $t$ |
| $\Rightarrow t^2 - t - 2 = 0 \Rightarrow t = 2, -1 \Rightarrow \left(2c, \frac{c}{2}\right)$ or $(-c, -c)$ | dM1 | Solves quadratic in $t$ |
| $(-c,\ -c)$ and $\left(2c,\ \frac{c}{2}\right)$ | A1 | Correct coordinates for both points |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Midpoint} = \left(\frac{-c+2c}{2},\ \frac{-c + \frac{c}{2}}{2}\right) = ...$ | M1 | Find midpoint of their $P$ and $Q$, provided not symmetric in $y$-axis |
| $x = \frac{c}{2}$ and $y = -\frac{c}{4} \Rightarrow xy = \frac{c}{2} \times -\frac{c}{4}$ leading to $xy = -\frac{c^2}{8}$ | A1cso | Uses correct midpoint coordinates to show $xy = -\frac{c^2}{8}$; must be an equation, no contrary statements |

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Show LaTeX source

\begin{enumerate}
  \item The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$ where $c$ is a positive constant.
\end{enumerate}

The line $l$ has equation $x - 2 y = c$\\
The points $P$ and $Q$ are the points of intersection of $H$ and $l$\\
(a) Determine, in terms of $c$, the coordinates of $P$ and the coordinates of $Q$

The point $R$ is the midpoint of $P Q$\\
(b) Show that, as $C$ varies, the coordinates of $R$ satisfy the equation

$$x y = - \frac { c ^ { 2 } } { a }$$

where $a$ is a constant to be determined.

\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q3 [5]}}

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