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

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.

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

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

$\left(3\cosh \theta \right)x-\left(4\sinh \theta \right)y=12$

Answer:

The tangent has the equation $y-y_1=m_T\left(x-x_1\right)$

Method 1

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

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

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

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

Method 2

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

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

After either method, substitute $m_T=\begin{array}{rcl}& & \frac{3\cosh \theta }{4\sinh \theta }\end{array}$, $x_1=4\cosh \theta$ and $y_1=3\sinh \theta$ into $y-y_1=m_T\left(x-x_1\right)$

$y-3\sinh \theta =\begin{array}{rcl}& & \frac{3\cosh \theta }{4\sinh \theta }\end{array}\left(x-4\cosh \theta \right)$

Rearrange into the form given in the question

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

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

$\left(3\cosh \theta \right)x-\left(4\sinh \theta \right)y=12$

Prove that, if $y=mx+c$ 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 $y=mx+c$ into the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

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 $x$

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

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

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

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

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

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

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

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

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

This rearranges to the answer

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

Show that the normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $P\left(a\sec \theta ,b\tan \theta \right)$ has the equation

$\left(a\sin \theta \right)x+by=\left(a^2+b^2\right)\tan \theta$

Answer:

The normal has the equation $y-y_1=m_N\left(x-x_1\right)$ 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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

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

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

Method 2

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

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

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

$m_N=-\frac{a\tan \theta }{b\sec \theta }$

Substitute $m_N=-\frac{a\tan \theta }{b\sec \theta }$, $x_1=a\sec \theta$ and $y_1=b\tan \theta$ into $y-y_1=m_N\left(x-x_1\right)$

$y-b\tan \theta =\begin{array}{rcl}& & -\frac{a\tan \theta }{b\sec \theta }\end{array}\left(x-a\sec \theta \right)$

Rearrange into the form given in the question

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

Simplify $\frac{\tan \theta }{\sec \theta }$

$\begin{array}{rcl}\frac{\tan \theta }{\sec \theta }& \equiv & \frac{\sin \theta }{\cos \theta }÷\frac{1}{\cos \theta }\\ & \equiv & \frac{\sin \theta }{\cancel{\cos \theta }}\times \frac{\cancel{\cos \theta }}{1}\\ & \equiv & \sin \theta \end{array}$

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

$\left(a\sin \theta \right)x+by=\left(a^2+b^2\right)\tan \theta$