A LevelFurther MathsEdexcelFurther Pure 1Revision NotesCoordinate SystemsTangent & Normals to Ellipses, Parabolas & HyperbolasTangents & Normals to Rectangular Hyperbolas

Tangents & Normals to Rectangular Hyperbolas (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Tangents & normals to rectangular hyperbolas

What is a tangent or normal to a rectangular hyperbola at a general point?

The position of the general point $P (c t, \frac{c}{t})$ on the rectangular hyperbola $x y = c^{2}$ depends on $t$
It is possible to calculate equations of tangents and normals at $P (c t, \frac{c}{t})$
- where the coefficients are in terms of $t$
  - i.e. as $P$ varies, the equations vary

Graph of rectangular hyperbola xy = c^2 with axes, point P(ct, c/t), and tangent and normal lines labelled at P.

In general
- at the point $P (c t, \frac{c}{t})$ on the rectangular hyperbola $x y = c^{2}$
  - $x + t^{2} y = 2 c t$ is the tangent
  - $t^{3} x - t y = c (t^{4} - 1)$ is the normal

Examiner Tips and Tricks

You are not expected to remember the general formulae for tangents and normals, but you are expected to be able to work them out using the steps below.

How do I find the equation of a tangent to a rectangular hyperbola?

To find the equation of the tangent to the rectangular hyperbola $x y = c^{2}$ at the general point $P (c t, \frac{c}{t})$ :
STEP 1
Find the gradient $m_{T}$ of the tangent at $P (c t, \frac{c}{t})$ in terms of $t$
- either by implicit differentiation of $x y = c^{2}$ to find $\frac{d y}{d x}$
  - then substituting $x = c t$ and $y = \frac{c}{t}$ into the result
- or by parametric differentiation of $x = c t$ and $y = \frac{c}{t}$
  - using $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}$
- or by making y the subject of $x y = c^{2}$ and finding $\frac{d y}{d x}$
  - i.e. $y = \frac{c^{2}}{x} = c^{2} x^{- 1}$ then differentiate
STEP 2
Substitute into the equation of a straight line $y - y_{1} = m_{T} (x - x_{1})$ the following:
- $m_{T}$ in terms of $t$
- $x_{1} = c t$
- $y_{1} = \frac{c}{t}$
- and simplify

Worked Example

Show that the tangent to the rectangular hyperbola $x y = 25$ at the point $P (5 t, \frac{5}{t})$ has the equation

$x + t^{2} y = 10 t$

Answer:

The tangent has the equation $y - y_{1} = m_{T} (x - x_{1})$

Method 1

Use implicit differentiation to differentiate $x y = 25$

$\begin{array}{rcl} 1 \times y + x \times \frac{d y}{d x} & = & 0 \\ y + x \frac{d y}{d x} & = & 0 \end{array}$

Substitute $x = 5 t$ and $y = \frac{5}{t}$ into the result and rearrange for $\frac{d y}{d x}$

$\begin{array}{rcl} (\frac{5}{t}) + (5 t) \frac{d y}{d x} & = & 0 \\ 5 t \frac{d y}{d x} & = & - \frac{5}{t} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

Method 2

Use parametric differentiation to find $\frac{d y}{d x}$ from $x = 5 t$ and $y = \frac{5}{t}$

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} & = & \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} \\ \frac{d y}{d x} & = & \frac{(- \frac{5}{t^{2}})}{5} \\ \frac{d y}{d x} & = & - \frac{5}{t^{2}} \div 5 \\ \frac{d y}{d x} & = & - \frac{5}{t^{2}} \times \frac{1}{5} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

Method 3

Make $y$ the subject of $x y = 25$

$y = \frac{25}{x} = 25 x^{- 1}$

Find $\frac{d y}{d x}$ using standard differentiation

$\begin{array}{rcl} \frac{d y}{d x} & = & - 25 x^{- 2} \\ \frac{d y}{d x} & = & - \frac{25}{x^{2}} \end{array}$

Substitute in $x = 5 t$ and simplify

$\begin{array}{rcl} \frac{d y}{d x} & = & - \frac{25}{{(5 t)}^{2}} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

After any of the methods above, substitute $m_{T} = \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{1}{t^{2}} \end{array}$ , $x_{1} = 5 t$ and $y_{1} = \frac{5}{t}$ into $y - y_{1} = m_{T} (x - x_{1})$

$y - \frac{5}{t} = - \frac{1}{t^{2}} (x - 5 t)$

Rearrange into the form given in the question

$\begin{array}{rcl} t^{2} y - 5 t & = & - (x - 5 t) \\ t^{2} y - 5 t & = & - x + 5 t \\ x + t^{2} y & = & 5 t + 5 t \end{array}$

This simplifies to the final answer

$x + t^{2} y = 10 t$

How do I find the equation of a normal to a rectangular hyperbola?

To find the equation of the normal to the rectangular hyperbola $x y = c^{2}$ at the general point $P (c t, \frac{c}{t})$ :
- follow the previous steps for finding the equation of a tangent
  - but use $y - y_{1} = m_{N} (x - x_{1})$ as the equation of the normal
  - where $m_{N} = - \frac{1}{m_{T}}$ is the negative reciprocal of the tangent gradient

Worked Example

Show that the normal to the rectangular hyperbola $x y = c^{2}$ at the point $P (c t, \frac{c}{t})$ has the equation

$t^{3} x - t y = c (t^{4} - 1)$

Answer:

The normal has the equation $y - y_{1} = m_{N} (x - x_{1})$ where the normal gradient is the negative reciprocal of the tangent gradient, $m_{N} = - \frac{1}{m_{T}}$

Method 1

Use implicit differentiation to differentiate $x y = c^{2}$

$\begin{array}{rcl} 1 \times y + x \times \frac{d y}{d x} & = & 0 \\ y + x \frac{d y}{d x} & = & 0 \end{array}$

Substitute $x = c t$ and $y = \frac{c}{t}$ into the result and rearrange for $\frac{d y}{d x}$ (the gradient of the tangent)

$\begin{array}{rcl} (\frac{c}{t}) + (c t) \frac{d y}{d x} & = & 0 \\ c t \frac{d y}{d x} & = & - \frac{c}{t} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

Method 2

Use parametric differentiation to find $\frac{d y}{d x}$ (the gradient of the tangent) from $x = c t$ and $y = \frac{c}{t}$

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} & = & \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} \\ \frac{d y}{d x} & = & \frac{(- \frac{c}{t^{2}})}{c} \\ \frac{d y}{d x} & = & - \frac{c}{t^{2}} \div c \\ \frac{d y}{d x} & = & - \frac{c}{t^{2}} \times \frac{1}{c} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

Method 3

Make $y$ the subject of $x y = c^{2}$

$y = \frac{c^{2}}{x} = c^{2} x^{- 1}$

Find $\frac{d y}{d x}$ using standard differentiation

$\begin{array}{rcl} \frac{d y}{d x} & = & - c^{2} x^{- 2} \\ \frac{d y}{d x} & = & - \frac{c^{2}}{x^{2}} \end{array}$

Substitute in $x = c t$ and simplify

$\begin{array}{rcl} \frac{d y}{d x} & = & - \frac{c^{2}}{{(c t)}^{2}} \\ \frac{d y}{d x} & = & - \frac{1}{t^{2}} \end{array}$

After any of the methods above, convert the tangent gradient into the normal gradient (e.g. find the negative reciprocal, or use $m_{N} = - \frac{1}{m_{T}}$ )

$m_{N} = t^{2}$

Substitute $m_{N} = t^{2}$ , $x_{1} = c t$ and $y_{1} = \frac{c}{t}$ into $y - y_{1} = m_{N} (x - x_{1})$

$y - \frac{c}{t} = t^{2} (x - c t)$

Rearrange into the form given in the question

$\begin{array}{rcl} t y - c & = & t^{3} (x - c t) \\ t y - c & = & t^{3} x - c t^{4} \\ c t^{4} - c & = & t^{3} x - t y \end{array}$

Factorise out $\begin{array}{rcl} c \end{array}$ to get the final answer

$t^{3} x - t y = c (t^{4} - 1)$

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent