Challenging +1.3 This is a multi-part Further Maths question on hyperbolas requiring implicit differentiation, parametric tangent equations, coordinate geometry, and locus finding. While it involves several steps and Further Maths content (hyperbolic functions, conic sections), each part follows standard techniques: (a) is routine implicit differentiation with hyperbolic identities, (b-c) are straightforward coordinate calculations, and (d) requires algebraic manipulation to eliminate the parameter. The question is methodical rather than requiring novel insight, making it moderately above average difficulty but not exceptionally challenging for FP3 students.
The hyperbola \(H\) has equation
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
Use calculus to show that the equation of the tangent to \(H\) at the point \((a\cosh\theta, b\sinh\theta)\) may be written in the form
$$xb\cosh\theta - ya\sinh\theta = ab$$
[4]
The line \(l_1\) is the tangent to \(H\) at the point \((a\cosh\theta, b\sinh\theta)\), \(\theta \neq 0\).
Given that \(l_1\) meets the \(x\)-axis at the point \(P\),
find, in terms of \(a\) and \(\theta\), the coordinates of \(P\).
[2]
The line \(l_2\) is the tangent to \(H\) at the point \((a, 0)\).
Given that \(l_1\) and \(l_2\) meet at the point \(Q\),
find, in terms of \(a\), \(b\) and \(\theta\), the coordinates of \(Q\).
[2]
Show that, as \(\theta\) varies, the locus of the mid-point of \(PQ\) has equation
$$x(4y^2 + b^2) = ab^2$$
[6]
Eliminate \(\sqrt{}\) and forms equation in x and y
Answer
Marks
Guidance
Simplifies to give required equation
1M1 A1ft, 2M1, 3M1, 4M1, A1cso
(6 marks)
Notes
Question 8(a):
- 1M1: Finding gradient in terms of \(\theta\)
- 1A1: CAO
- 2M1: Finding equation of tangent
- 2A1: CSO (answer given) look for \(\pm(\cosh^2\theta - \sinh^2\theta)\)
Question 8(b):
- M1: Putting \(y=0\) into their tangent
- A1ft: P found, ft for their tangent o.e.
Question 8(c):
- M1: Putting \(x=a\) into their tangent
- A1: CAO Q found o.e.
Question 8(d) - For Alt 1 and 2:
- 1M1: Finding expressions, in terms of \(\sinh\theta\) and \(\cosh\theta\) but must be adding
- 1A1: Ft on their P and Q
- 2M1: Finding \(4y^2+b^2\)
- 3M1: Simplified, factorised, maximum of 2 terms per bracket
- 4M1: Finding \(x(4y^2+b^2)\), completely factorised, maximum of 2 terms per bracket
- 2A1: CSO
Question 8(d) - For Alts 3, 4 and 5:
- 1M1: Finding expressions, in terms of \(\sinh\theta\) and \(\cosh\theta\) but must be adding
- 1A1: Ft on their P and Q
- 2M1: Getting \(\cosh\theta\) in terms of \(x\)
- 3M1: \(y\) or \(y^2\) in terms of \(\cosh\theta\) or \(\sinh\theta\) in terms of \(x\) and \(y\)
- 4M1: Getting equation in terms of \(x\) and \(y\) only. No square roots.
- 2A1: CSO
## Part (a)
**Answer/Working:** Uses $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cosh\theta}{a\sinh\theta}$ or $\frac{2x}{a^2} - \frac{2yy'}{b^2} = 0 \to y' = \frac{xb^2}{ya^2} - \frac{b\cosh\theta}{a\sinh\theta}$
So $y - b\sinh\theta = \frac{b\cosh\theta}{a\sinh\theta}(x - a\cosh\theta)$
$\therefore ab(\cosh^2\theta - \sinh^2\theta) = xb\cosh\theta - ya\sinh\theta$ and as $(\cosh^2\theta - \sinh^2\theta) = 1$
$xb\cosh\theta - ya\sinh\theta = ab$ * | **M1 A1, M1, A1cso** | (4 marks)
## Part (b)
**Answer/Working:** $P$ is the point $\left(\frac{a}{\cosh\theta}, 0\right)$ | **M1 A1** | (2 marks)
## Part (c)
**Answer/Working:** $l_2$ has equation $x = a$ and meets $l_1$ at Q$(a, \frac{b(\cosh\theta-1)}{\sinh\theta})$ | **M1 A1** | (2 marks)
## Part (d) - Alt 1
**Answer/Working:** The mid point of $PQ$ is given by $X = \frac{a(\cosh\theta+1)}{2\cosh\theta}$, $Y = \frac{b(\cosh\theta-1)}{2\sinh\theta}$
$4Y^2 + b^2 = b^2\left(\frac{\cosh^2\theta+1-2\cosh\theta+\sinh^2\theta}{\sinh^2\theta}\right)$ | **1M1 A1ft, 2M1**
$= b^2\left(\frac{2\cosh^2\theta-2\cosh\theta}{\sinh^2\theta}\right)$ | **3M1**
$X(4Y^2 + b^2) = ab^2\left(\frac{(\cosh\theta+1)(\cosh\theta-1)2\cosh\theta}{2\cosh\theta\sinh^2\theta}\right)$ | **4M1**
Simplify fraction by using $\cosh^2\theta - \sinh^2\theta = 1$ to give $x(4y^2+b^2) = ab^2$ * | **A1cso** | (6 marks)
## Part (d) - Alt 2
**Answer/Working:** First line of solution as before
$4Y^2 + b^2 = b^2\left(\coth^2\theta + \text{cosech}^2\theta - 2\coth\theta\text{cosech}\theta + 1\right)$ | **1M1A1ft, 2M1**
$= b^2\left(2\coth^2\theta - 2\coth\theta\text{cosech}\theta\right)$ | **3M1**
$X(4Y^2 + b^2) = ab^2(\coth\theta(\coth\theta - \text{cosech}\theta)(1+\text{sech}\theta))$ | **4M1**
Simplify expansion by using $\coth^2\theta - \text{cosech}^2\theta = 1$ to give $x(4y^2+b^2) = ab^2$ * | **A1cso** | (6 marks)
---
## Part (d) - Alt 3
**Answer/Working:** $X = \frac{a(\cosh\theta+1)}{2\cosh\theta}$, $Y = \frac{b(\cosh\theta-1)}{2\sinh\theta}$
$\cosh\theta = \frac{a}{2x-a}$
$\sinh\theta = \frac{b(\cosh\theta-1)}{2y} = \frac{b(a-x)}{(2x-a)y}$
$\left(\frac{a}{2x-a}\right)^2 - \left(\frac{b(a-x)}{(2x-a)y}\right)^2 = 1$ | **1M1 A1ft, 2M1, 3M1, 4M1**
Simplifies to give required equation | **A1cso** | (6 marks)
---
## Part (d) - Alt 4
**Answer/Working:** $X = \frac{a(\cosh\theta+1)}{2\cosh\theta}$, $Y = \frac{b(\cosh\theta-1)}{2\sinh\theta}$
$\cosh\theta = \frac{a}{2x-a}$
$y^2 = \frac{b^2(\cosh\theta-1)^2}{4(\cosh^2\theta-1)} = \frac{b^2(\cosh\theta-1)}{4(\cosh\theta+1)}$ | **1M1 A1ft, 2M1, 3M1**
$y^2 = \frac{b^2\left(\frac{2a-2x}{2x-a}\right)}{4\left(\frac{2x}{2x-a}\right)}$ o.e | **4M1**
Simplifies to give required equation | **A1 cso** | (6 marks)
---
## Part (d) - Alt 5
**Answer/Working:** $X = \frac{a(\cosh\theta+1)}{2\cosh\theta}$, $Y = \frac{b(\cosh\theta-1)}{2\sinh\theta}$
$\cosh\theta = \frac{a}{2x-a}$
$y = \left(\frac{b(\cosh\theta-1)}{2\sinh\theta}\right) = \left(\frac{b(\cosh\theta-1)}{2\sqrt{\cosh^2\theta-1}}\right)$
Eliminate $\sqrt{}$ and forms equation in x and y
Simplifies to give required equation | **1M1 A1ft, 2M1, 3M1, 4M1, A1cso** | (6 marks)
---
# Notes
**Question 8(a):**
- 1M1: Finding gradient in terms of $\theta$
- 1A1: CAO
- 2M1: Finding equation of tangent
- 2A1: CSO (answer given) look for $\pm(\cosh^2\theta - \sinh^2\theta)$
**Question 8(b):**
- M1: Putting $y=0$ into their tangent
- A1ft: P found, ft for their tangent o.e.
**Question 8(c):**
- M1: Putting $x=a$ into their tangent
- A1: CAO Q found o.e.
**Question 8(d) - For Alt 1 and 2:**
- 1M1: Finding expressions, in terms of $\sinh\theta$ and $\cosh\theta$ but must be adding
- 1A1: Ft on their P and Q
- 2M1: Finding $4y^2+b^2$
- 3M1: Simplified, factorised, maximum of 2 terms per bracket
- 4M1: Finding $x(4y^2+b^2)$, completely factorised, maximum of 2 terms per bracket
- 2A1: CSO
**Question 8(d) - For Alts 3, 4 and 5:**
- 1M1: Finding expressions, in terms of $\sinh\theta$ and $\cosh\theta$ but must be adding
- 1A1: Ft on their P and Q
- 2M1: Getting $\cosh\theta$ in terms of $x$
- 3M1: $y$ or $y^2$ in terms of $\cosh\theta$ or $\sinh\theta$ in terms of $x$ and $y$
- 4M1: Getting equation in terms of $x$ and $y$ only. No square roots.
- 2A1: CSO
The hyperbola $H$ has equation
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that the equation of the tangent to $H$ at the point $(a\cosh\theta, b\sinh\theta)$ may be written in the form
$$xb\cosh\theta - ya\sinh\theta = ab$$
[4]
The line $l_1$ is the tangent to $H$ at the point $(a\cosh\theta, b\sinh\theta)$, $\theta \neq 0$.
Given that $l_1$ meets the $x$-axis at the point $P$,
\item find, in terms of $a$ and $\theta$, the coordinates of $P$.
[2]
The line $l_2$ is the tangent to $H$ at the point $(a, 0)$.
Given that $l_1$ and $l_2$ meet at the point $Q$,
\item find, in terms of $a$, $b$ and $\theta$, the coordinates of $Q$.
[2]
\item Show that, as $\theta$ varies, the locus of the mid-point of $PQ$ has equation
$$x(4y^2 + b^2) = ab^2$$
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q8 [14]}}