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

Exam code: 9FM0

Author

Mark Curtis

Last updated

7 January 2026

Tangents & normals to hyperbolas

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

The position of the general point $P (a \cosh θ, b \sinh θ)$ or $P (a \sec θ, b \tan θ)$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ depends on $θ$
It is possible to calculate equations of tangents and normals at $P (a \cosh θ, b \sinh θ)$ or $P (a \sec θ, b \tan θ)$
- where the coefficients are in terms of $θ$
  - i.e. as $P$ varies, the equations vary

The hyperbola x^2/a^2-y^2/b^2=1 shown with a point P with coordinates (a cosh theta, b sinh theta) or P(a sec theta. b tan theta). The tangent and the normal at P are drawn on the graph.

In general
- at the point $P (a \cosh θ, b \sinh θ)$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
  - $(b \cosh θ) x - (a \sinh θ) y = a b$ is the tangent
  - $(a \sinh θ) x + (b \cosh θ) y = (a^{2} + b^{2}) \sinh θ \cosh θ$ is the normal
- at the point $P (a \sec θ, b \tan θ)$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
  - $(b \sec θ) x - (a \tan θ) y = a b$ is the tangent
  - $(a \sin θ) x + b y = (a^{2} + b^{2}) \tan θ$ is the normal
Be careful with infinite gradients at the vertices
- e.g. the equation of the tangent at $(a, 0)$ is $x = a$

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 hyperbola?

To find the equation of the tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at the general point $P (a \cosh θ, b \sinh θ)$ or $P (a \sec θ, b \tan θ)$ :
STEP 1
Find the gradient $m_{T}$ of the tangent at $P (a \cosh θ, b \sinh θ)$ or $P (a \sec θ, b \tan θ)$ in terms of $θ$
- either by implicit differentiation of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to find $\frac{d y}{d x}$
  - then substituting $x = a \cosh θ$ and $y = b \sinh θ$ (or $x = a \sec θ$ and $y = b \tan θ$ ) into the result
- or by parametric differentiation of $x = a \cosh θ$ and $y = b \sinh θ$ (or $x = a \sec θ$ and $y = b \tan θ$ )
  - using $\frac{d y}{d x} = \frac{d y}{d θ} \times \frac{d θ}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})}$
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 $θ$
- $x_{1} = a \cosh θ$ (or $x_{1} = a \sec θ$ )
- $y_{1} = b \sinh θ$ (or $y_{1} = b \tan θ$ )
- and simplify using hyperbolic or trig identities

Examiner Tips and Tricks

It is possible to make $y$ the subject of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to find $\frac{d y}{d x}$ , i.e. $y = \pm \sqrt{b^{2} (\frac{x^{2}}{a^{2}} - 1)}$ , but differentiating this is more messy than implicit or parametric differentiation!

Worked Example

Show that the tangent to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at the point $P (4 \cosh θ, 3 \sinh θ)$ has the equation

$(3 \cosh θ) x - (4 \sinh θ) y = 12$

Answer:

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

Method 1

Use implicit differentiation to differentiate $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

$\frac{2 x}{16} - \frac{2 y}{9} \frac{d y}{d x} = 0$

Substitute $x = 4 \cosh θ$ and $y = 3 \sinh θ$ into the result and rearrange for $\frac{d y}{d x}$

$\begin{array}{rcl} \frac{2 (4 \cosh θ)}{16} - \frac{2 (3 \sinh θ)}{9} \frac{d y}{d x} & = & 0 \\ \frac{2 (3 \sinh θ)}{9} \frac{d y}{d x} & = & \frac{2 (4 \cosh θ)}{16} \\ \frac{\sinh θ}{3} \frac{d y}{d x} & = & \frac{\cosh θ}{4} \\ \frac{d y}{d x} & = & \frac{3 \cosh θ}{4 \sinh θ} \end{array}$

Method 2

Use parametric differentiation to find $\frac{d y}{d x}$ from $x = 4 \cosh θ$ and $y = 3 \sinh θ$

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d θ} \times \frac{d θ}{d x} \\ \frac{d y}{d x} & = & \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} \\ \frac{d y}{d x} & = & \frac{3 \cosh θ}{4 \sinh θ} \end{array}$

After either method, substitute $m_{T} = \begin{array}{rcl} \frac{3 \cosh θ}{4 \sinh θ} \end{array}$ , $x_{1} = 4 \cosh θ$ and $y_{1} = 3 \sinh θ$ into $y - y_{1} = m_{T} (x - x_{1})$

$y - 3 \sinh θ = \begin{array}{rcl} \frac{3 \cosh θ}{4 \sinh θ} \end{array} (x - 4 \cosh θ)$

Rearrange into the form given in the question

$\begin{array}{rcl} (4 \sinh θ) y - 12 \sinh^{2} θ & = & (3 \cosh θ) x - 12 \cosh^{2} θ \\ 12 (\cosh^{2} θ - \sinh^{2} θ) & = & (3 \cosh θ) x - (4 \sinh θ) y \end{array}$

Use that $\cosh^{2} θ - \sinh^{2} θ \equiv 1$ to get the final answer

$(3 \cosh θ) x - (4 \sinh θ) y = 12$

What is the tangent condition for a hyperbola?

The condition for a straight line $y = m x + c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is that the gradient $m$ and y-intercept $c$ of the straight line must satisfy
- $a^{2} m^{2} - b^{2} = c^{2}$
You need to know how to prove this condition
- by solving $y = m x + c$ and $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ simultaneously
- and forcing the discriminant to be zero
  - See the worked example below

Worked Example

Prove that, if $y = m x + c$ is tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , then $a^{2} m^{2} - b^{2} = c^{2}$ .

Answer:

First substitute $y = m x + c$ into the equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{x^{2}}{a^{2}} - \frac{{(m x + c)}^{2}}{b^{2}} = 1$

It helps to

multiply both sides by $a^{2} b^{2}$
add the bracketed term to the right-hand side (to make expanding easier)
then expand and rearrange into a three-term quadratic in $x$

$\begin{array}{rcl} b^{2} x^{2} - a^{2} {(m x + c)}^{2} & = & a^{2} b^{2} \\ b^{2} x^{2} & = & a^{2} b^{2} + a^{2} {(m x + c)}^{2} \\ b^{2} x^{2} & = & a^{2} b^{2} + a^{2} (m^{2} x^{2} + 2 m c x + c^{2}) \\ 0 & = & (a^{2} m^{2} - b^{2}) x^{2} + 2 a^{2} m c x + a^{2} (b^{2} + c^{2}) \end{array}$

The solutions to this equation are the $x$ -intercepts of the points of intersection

Force the discriminant to be zero, as a tangent only touches the hyperbola once

${(2 a^{2} m c)}^{2} - 4 (a^{2} m^{2} - b^{2}) \times a^{2} (b^{2} + c^{2}) = 0$

It helps to move the second half to the other side, to make expanding easier

${(2 a^{2} m c)}^{2} = 4 (a^{2} m^{2} - b^{2}) \times a^{2} (b^{2} + c^{2})$

Factorise out $4 a^{2}$ from both sides, cancel, then expand the brackets and cancel any common terms on both sides

$\begin{array}{rcl} 4 a^{4} m^{2} c^{2} & = & 4 a^{2} (a^{2} m^{2} - b^{2}) (b^{2} + c^{2}) \\ 4 a^{2} a^{2} m^{2} c^{2} & = & 4 a^{2} (a^{2} m^{2} - b^{2}) (b^{2} + c^{2}) \\ a^{2} m^{2} c^{2} & = & a^{2} m^{2} b^{2} + a^{2} m^{2} c^{2} - b^{4} - b^{2} c^{2} \\ 0 & = & a^{2} m^{2} b^{2} - b^{4} - b^{2} c^{2} \end{array}$

Factorise out $b^{2}$ and cancel (as $b \neq 0$ in $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ )

$\begin{array}{rcl} 0 & = & b^{2} (a^{2} m^{2} - b^{2} - c^{2}) \\ 0 & = & a^{2} m^{2} - b^{2} - c^{2} \end{array}$

This rearranges to the answer

$a^{2} m^{2} - b^{2} = c^{2}$

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

To find the equation of the normal to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at the general point $P (a \cosh θ, b \sinh θ)$ or $P (a \sec θ, b \tan θ)$ :
- 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 hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at the point $P (a \sec θ, b \tan θ)$ has the equation

$(a \sin θ) x + b y = (a^{2} + b^{2}) \tan θ$

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

$\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0$

Substitute $x = a \sec θ$ and $y = b \tan θ$ into the result and rearrange for $\frac{d y}{d x}$ (the gradient of the tangent)

$\begin{array}{rcl} \frac{2 (a \sec θ)}{a^{2}} - \frac{2 (b \tan θ)}{b^{2}} \frac{d y}{d x} & = & 0 \\ - \frac{2 (b \tan θ)}{b^{2}} \frac{d y}{d x} & = & - \frac{2 (a \sec θ)}{a^{2}} \\ \frac{\tan θ}{b} \frac{d y}{d x} & = & \frac{\sec θ}{a} \\ \frac{d y}{d x} & = & \frac{b \sec θ}{a \tan θ} \end{array}$

Method 2

Use parametric differentiation to find $\frac{d y}{d x}$ (the gradient of the tangent) from $x = a \sec θ$ and $y = b \tan θ$

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d θ} \times \frac{d θ}{d x} \\ \frac{d y}{d x} & = & \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} \\ \frac{d y}{d x} & = & \frac{b \sec^{2} θ}{a \sec θ \tan θ} \\ \frac{d y}{d x} & = & \frac{b \sec θ}{a \tan θ} \end{array}$

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

$m_{N} = - \frac{a \tan θ}{b \sec θ}$

Substitute $m_{N} = - \frac{a \tan θ}{b \sec θ}$ , $x_{1} = a \sec θ$ and $y_{1} = b \tan θ$ into $y - y_{1} = m_{N} (x - x_{1})$

$y - b \tan θ = \begin{array}{rcl} - \frac{a \tan θ}{b \sec θ} \end{array} (x - a \sec θ)$

Rearrange into the form given in the question

$\begin{array}{rcl} y - b \tan θ & = & - (\frac{a \tan θ}{b \sec θ}) x + \frac{a^{2}}{b} \tan θ \\ b y - b^{2} \tan θ & = & - (\frac{a \tan θ}{\sec θ}) x + a^{2} \tan θ \\ (\frac{a \tan θ}{\sec θ}) x + b y & = & (a^{2} + b^{2}) \tan θ \end{array}$

Simplify $\frac{\tan θ}{\sec θ}$

$\begin{array}{rcl} \frac{\tan θ}{\sec θ} & \equiv & \frac{\sin θ}{\cos θ} \div \frac{1}{\cos θ} \\ \equiv & \frac{\sin θ}{\cos θ} \times \frac{\cos θ}{1} \\ \equiv & \sin θ \end{array}$

The answer can now be written in the form given in the question

$(a \sin θ) x + b y = (a^{2} + b^{2}) \tan θ$

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

Tangents & normals to hyperbolas

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

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

What is the tangent condition for a hyperbola?

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

Unlock more, it's free!

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

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Author: Mark Curtis