Question 8 - A-Level Maths

Edexcel FP1 2010 June — Question 8 11 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Rectangular hyperbola normal re-intersection
Difficulty	Challenging +1.2 This is a standard FP1 rectangular hyperbola question requiring implicit differentiation to find the normal, then solving a cubic equation when the normal re-intersects the curve. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are routine for FP1 students: finding coordinates from xy=c², differentiating implicitly, writing the normal equation, and substituting back. The algebra is straightforward with no conceptual surprises, making it a typical textbook exercise rather than a problem requiring novel insight.
Spec	1.02q Use intersection points: of graphs to solve equations 1.07m Tangents and normals: gradient and equations 1.07s Parametric and implicit differentiation

8. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where c is a positive constant. The point $A$ on $H$ has $x$-coordinate $3 c$.

Write down the $y$-coordinate of $A$.
Show that an equation of the normal to $H$ at $A$ is $$3 y = 27 x - 80 c$$ The normal to $H$ at $A$ meets $H$ again at the point $B$.
Find, in terms of $c$, the coordinates of $B$.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $\frac{c}{3}$	B1	(1 mark)
(b) $y = \frac{c^2}{x} \Rightarrow \frac{dy}{dx} = -c^2 x^{-2}$	B1
Or: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$ or $\dot{x} = c, \dot{y} = -\frac{c}{t^2}$ so $\frac{dy}{dx} = -\frac{1}{t^2}$
and at $A$: $\frac{dy}{dx} = -\frac{c^2}{(3c)^2} = -\frac{1}{9}$, so gradient of normal is $9$	M1 A1	M1: Substitutes values and uses negative reciprocal (needs to follow calculus). A1: $9$ cao (needs to follow calculus)
Either: $y - \frac{c}{3} = 9(x - 3c)$ or $y = 9x + k$ and use $x = 3c, y = \frac{c}{3}$	M1
$\Rightarrow 3y = 27x - 80c$	A1	(5 marks)

Guidance:

(b) B1: Any valid method of differentiation but must get to correct expression for $\frac{dy}{dx}$

M1: Substitutes values and uses negative reciprocal (needs to follow calculus)

A1: $9$ cao (needs to follow calculus)

M1: Finds equation of line through $A$ with any gradient (other than $0$ and $\infty$)

A1: Correct work throughout – obtaining printed answer.

Answer	Marks	Guidance
(c) $\frac{c^2}{x} = \frac{27x - 80c}{3}$	M1
$3c^2 = 27x^2 - 80cx$	A1
$(x - 3c)(27x + c) = 0$ so $x = -\frac{c}{27}, y = -27c$	M1, A1, A1
Or: $\frac{c^2}{y} = \frac{3y + 80c}{27}$, $27c^2 = 3y^2 + 80cy$, $(y + 27c)(3y - c) = 0$ so $x = -\frac{c}{27}, y = -27c$
Or: $\frac{3c}{t} = \frac{27et - 80c}{3}$, $3c = 27et^2 - 80ct$, $(t - 3)(27t + 1) = 0$ so $t = -\frac{1}{27}$ (and so) $x = -\frac{c}{27}, y = -27c$		(5 marks)

Total: 11 marks

Guidance:

(c) M1: Obtains equation in one variable ($x, y$ or $t$). A1: Writes as correct three-term quadratic (any equivalent form). M1: Attempts to solve three-term quadratic to obtain $x = $ or $y = $ or $t = $. A1: $x$ coordinate. A1: $y$ coordinate. (cao but allow recovery following slips)

**(a)** $\frac{c}{3}$ | **B1** | (1 mark)

**(b)** $y = \frac{c^2}{x} \Rightarrow \frac{dy}{dx} = -c^2 x^{-2}$ | **B1** | 

Or: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$ or $\dot{x} = c, \dot{y} = -\frac{c}{t^2}$ so $\frac{dy}{dx} = -\frac{1}{t^2}$ | | 

and at $A$: $\frac{dy}{dx} = -\frac{c^2}{(3c)^2} = -\frac{1}{9}$, so gradient of normal is $9$ | **M1 A1** | M1: Substitutes values and uses negative reciprocal (needs to follow calculus). A1: $9$ cao (needs to follow calculus)

Either: $y - \frac{c}{3} = 9(x - 3c)$ or $y = 9x + k$ and use $x = 3c, y = \frac{c}{3}$ | **M1** | 

$\Rightarrow 3y = 27x - 80c$ | **A1** | (5 marks)

**Guidance:**

(b) B1: Any valid method of differentiation but must get to correct expression for $\frac{dy}{dx}$

M1: Substitutes values and uses negative reciprocal (needs to follow calculus)

A1: $9$ cao (needs to follow calculus)

M1: Finds equation of line through $A$ with any gradient (other than $0$ and $\infty$)

A1: Correct work throughout – obtaining printed answer.

**(c)** $\frac{c^2}{x} = \frac{27x - 80c}{3}$ | **M1** | 

$3c^2 = 27x^2 - 80cx$ | **A1** | 

$(x - 3c)(27x + c) = 0$ so $x = -\frac{c}{27}, y = -27c$ | **M1, A1, A1** | 

Or: $\frac{c^2}{y} = \frac{3y + 80c}{27}$, $27c^2 = 3y^2 + 80cy$, $(y + 27c)(3y - c) = 0$ so $x = -\frac{c}{27}, y = -27c$ | | 

Or: $\frac{3c}{t} = \frac{27et - 80c}{3}$, $3c = 27et^2 - 80ct$, $(t - 3)(27t + 1) = 0$ so $t = -\frac{1}{27}$ (and so) $x = -\frac{c}{27}, y = -27c$ | | (5 marks)

**Total: 11 marks**

**Guidance:**

(c) M1: Obtains equation in one variable ($x, y$ or $t$). A1: Writes as correct three-term quadratic (any equivalent form). M1: Attempts to solve three-term quadratic to obtain $x = $ or $y = $ or $t = $. A1: $x$ coordinate. A1: $y$ coordinate. (cao but allow recovery following slips)

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Show LaTeX source

8. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where c is a positive constant. The point $A$ on $H$ has $x$-coordinate $3 c$.
\begin{enumerate}[label=(\alph*)]
\item Write down the $y$-coordinate of $A$.
\item Show that an equation of the normal to $H$ at $A$ is

$$3 y = 27 x - 80 c$$

The normal to $H$ at $A$ meets $H$ again at the point $B$.
\item Find, in terms of $c$, the coordinates of $B$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP1 2010 Q8 [11]}}

This paper (9 questions)

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Q1 7 Q2 5 Q3 10 Q4 7 Q5 5 Q6 9 Q7 7 Q8 11 Q9 14

Edexcel FP1 2010 June — Question 8 11 marks

Guidance:

(b) B1: Any valid method of differentiation but must get to correct expression for \(\frac{dy}{dx}\)

M1: Substitutes values and uses negative reciprocal (needs to follow calculus)

A1: \(9\) cao (needs to follow calculus)

M1: Finds equation of line through \(A\) with any gradient (other than \(0\) and \(\infty\))

A1: Correct work throughout – obtaining printed answer.

Total: 11 marks

Guidance:

(c) M1: Obtains equation in one variable (\(x, y\) or \(t\)). A1: Writes as correct three-term quadratic (any equivalent form). M1: Attempts to solve three-term quadratic to obtain \(x = \) or \(y = \) or \(t = \). A1: \(x\) coordinate. A1: \(y\) coordinate. (cao but allow recovery following slips)

This paper (9 questions)

Answer	Marks	Guidance
(a) \(\frac{c}{3}\)	B1	(1 mark)
(b) \(y = \frac{c^2}{x} \Rightarrow \frac{dy}{dx} = -c^2 x^{-2}\)	B1
Or: \(y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}\) or \(\dot{x} = c, \dot{y} = -\frac{c}{t^2}\) so \(\frac{dy}{dx} = -\frac{1}{t^2}\)
and at \(A\): \(\frac{dy}{dx} = -\frac{c^2}{(3c)^2} = -\frac{1}{9}\), so gradient of normal is \(9\)	M1 A1	M1: Substitutes values and uses negative reciprocal (needs to follow calculus). A1: \(9\) cao (needs to follow calculus)
Either: \(y - \frac{c}{3} = 9(x - 3c)\) or \(y = 9x + k\) and use \(x = 3c, y = \frac{c}{3}\)	M1
\(\Rightarrow 3y = 27x - 80c\)	A1	(5 marks)

Answer	Marks	Guidance
(c) \(\frac{c^2}{x} = \frac{27x - 80c}{3}\)	M1
\(3c^2 = 27x^2 - 80cx\)	A1
\((x - 3c)(27x + c) = 0\) so \(x = -\frac{c}{27}, y = -27c\)	M1, A1, A1
Or: \(\frac{c^2}{y} = \frac{3y + 80c}{27}\), \(27c^2 = 3y^2 + 80cy\), \((y + 27c)(3y - c) = 0\) so \(x = -\frac{c}{27}, y = -27c\)
Or: \(\frac{3c}{t} = \frac{27et - 80c}{3}\), \(3c = 27et^2 - 80ct\), \((t - 3)(27t + 1) = 0\) so \(t = -\frac{1}{27}\) (and so) \(x = -\frac{c}{27}, y = -27c\)		(5 marks)