The ellipse and hyperbolaEdexcel International A Level Further Maths: Revision notes

Section 1

The ellipse: Cartesian and parametric forms

The ellipse with centre the origin has Cartesian equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ crossing the axes at $(\pm a,0)$ and $(0,\pm b)$ . Its parametric equations are $x=a\cos t$ , $y=b\sin t$ , $0\le t<2\pi$ . They satisfy the Cartesian equation because $\cos^2t+\sin^2t=1$ . The parameter $t$ is not the angle from the origin to the point, except for a circle. Example: $x=5\cos t$ , $y=3\sin t$ gives $\frac{x^2}{25}+\frac{y^2}{9}=1$ . At $t=\frac{\pi}{3}$ the point is $\left(\frac52,\frac{3\sqrt3}{2}\right)$ .

Key termsellipseparametric equationsparameter

Common mistake

Writing $\frac{x}{a^2}+\frac{y}{b^2}=1$ , or swapping $a$ and $b$ . The $x$ -denominator is $a^2$ , where $a=x$ -intercept.

Section 2

The hyperbola: Cartesian form and the sec/tan parametrisation

The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has two branches, crossing the $x$ -axis at $(\pm a,0)$ and approaching the asymptotes $y=\pm\frac bax$ . A parametrisation is $x=a\sec t,\qquad y=b\tan t,$ since $\sec^2t-\tan^2t=1$ (from $1+\tan^2t=\sec^2t$ ). As $t$ goes from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ the point traces the right branch, and for $\frac{\pi}{2}<t<\frac{3\pi}{2}$ the left branch. Example: $x=4\sec t$ , $y=3\tan t$ gives $\frac{x^2}{16}-\frac{y^2}{9}=1$ . At $t=\frac{\pi}{3}$ the point is $\left(8,3\sqrt3\right)$ .

Key termshyperbolabranchasymptote

Exam tip

For the Cartesian form from $\sec$ and $\tan$ , write $\sec t=\frac xa$ , $\tan t=\frac yb$ and use $1+\tan^2t=\sec^2t$ .

Section 3

The cosh and sinh parametrisation

A second parametrisation of the right branch uses hyperbolic functions: $x=a\cosh t,\qquad y=b\sinh t,\quad t\in\mathbb{R},$ because $\cosh^2t-\sinh^2t=1$ gives $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . Since $\cosh t\ge1$ this covers only $x\ge a$ , the right branch. The left branch is $x=-a\cosh t$ , $y=b\sinh t$ . Example: $x=3\cosh t$ , $y=4\sinh t$ gives $\frac{x^2}{9}-\frac{y^2}{16}=1$ . At $t=\ln2$ , $\cosh t=\frac54$ and $\sinh t=\frac34$ , so the point is $\left(\frac{15}{4},3\right)$ .

Key termscosh parametrisation

Common mistake

Using $x=a\sinh t$ , $y=b\cosh t$ . That gives $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ , a different hyperbola.

Section 4

Finding points and parameter values

To find the point for a given $t$ , substitute. To find $t$ for a given point, equate each coordinate: for the ellipse $x=5\cos t$ , $y=3\sin t$ and the point $\left(4,-\frac95\right)$ , $\cos t=\frac45$ and $\sin t=-\frac35$ , so $t$ is in the fourth quadrant and $t=2\pi-\tan^{-1}\frac34=5.64$ . For a hyperbola, switching between forms uses $x=a\sec t$ or $x=a\cosh u$ : for $P=\left(8,3\sqrt3\right)$ on $\frac{x^2}{16}-\frac{y^2}{9}=1$ , $4\cosh u=8$ gives $u=\ln\left(2+\sqrt3\right)$ .

Exam tip

Check both coordinates when finding $t$ , because two values of $t$ can share the same $x$ .

Section 5

Using the parametrisation

Parametric forms turn geometry into trigonometry. For the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ with $x=4\cos t$ , $y=2\sin t$ : $OP^2=16\cos^2t+4\sin^2t=4+12\cos^2t$ , which runs from $4$ to $16$ , so $OP$ ranges from $2$ to $4$ (the semi-minor and semi-major axes). To find where a line meets a conic, substitute the line into the Cartesian equation (or substitute the parametric coordinates into the line). If a line is parallel to an asymptote the $x^2$ terms cancel, leaving a linear equation and just one intersection: $3x-4y=12$ meets $\frac{x^2}{16}-\frac{y^2}{9}=1$ only at $(4,0)$ .

Exam tip

After substituting a line, if the $x^2$ terms vanish, check whether the line is parallel to an asymptote.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on The ellipse and hyperbola

The ellipse $E$ has parametric equations $x=5\cos t$ , $y=3\sin t$ , for $0\le t<2\pi$ .
Find the value of $t$ , for $0\le t<2\pi$ , at the point $\left(4,-\frac95\right)$ on $E$ . Give your answer to 3 significant figures.2 marks
The hyperbola $H$ has Cartesian equation $\frac{x^2}{9}-\frac{y^2}{16}=1$ .
Show that $x=3\cosh t$ , $y=4\sinh t$ satisfies the Cartesian equation of $H$ .2 marks
The ellipse $E$ has Cartesian equation $\frac{x^2}{16}+\frac{y^2}{4}=1$ .
Write down parametric equations for $E$ . Hence find the exact value of the parameter $t$ , for $0<t<\frac{\pi}{2}$ , at the point $\left(2\sqrt2,\sqrt2\right)$ on $E$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).