The equations and key features of ellipses and hyperbolas, including centre, vertices, axes and asymptotes, and the description of curves by parametric equations together with conversion between parametric and Cartesian forms

What this dot point is asking

VCAA wants you to recognise and sketch ellipses and hyperbolas from their standard equations, identifying centre, vertices, axis lengths and (for hyperbolas) asymptotes, and to describe curves parametrically, converting between parametric and Cartesian descriptions. Parametric forms reappear in calculus for arc length and motion, so fluency here pays off later.

Ellipses

The standard ellipse centred at $(h, k)$ is

The graph extends $a$ units left and right of the centre and $b$ units up and down. If $a > b$ the major axis is horizontal; if $b > a$ it is vertical. The vertices on the horizontal axis are $(h \pm a, k)$ and on the vertical axis $(h, k \pm b)$. When $a = b$ the ellipse is a circle of radius $a$.

Hyperbolas

The standard hyperbola opening horizontally is

It has two branches with vertices at $(h \pm a, k)$ and no points between the branches. As $|x| \to \infty$ the curve approaches the asymptotes

A quick way to draw the asymptotes is to sketch the rectangle of half-width $a$ and half-height $b$ centred at $(h, k)$; the asymptotes are the extended diagonals. The vertical hyperbola $\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1$ opens up and down with vertices at $(h, k \pm b)$.

Parametric curves

A parametric description gives both coordinates as functions of a parameter:

As $t$ runs over its allowed values the point $(x, y)$ traces a curve. To find the Cartesian equation, eliminate $t$: solve one equation for $t$ and substitute, or use an identity that links the two expressions.

The ellipse has the natural parametrisation $x = h + a\cos t$, $y = k + b\sin t$ for $t \in [0, 2\pi)$, because then $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = \cos^2 t + \sin^2 t = 1$. A hyperbola can be parametrised with $x = h + a\sec t$, $y = k + b\tan t$, using $\sec^2 t - \tan^2 t = 1$.

Examples in context

Example 1. $\frac{x^2}{25} - \frac{y^2}{9} = 1$ is a hyperbola with vertices $(\pm 5, 0)$ and asymptotes $y = \pm\frac{3}{5}x$.

Example 2. $x = t$, $y = t^2$ gives $y = x^2$, the standard parabola, parametrised left to right as $t$ increases.

Try this

Q1. State the centre and semi-axes of $\frac{(x + 1)^2}{16} + \frac{y^2}{9} = 1$. [2 marks]

• Cue. Centre $(-1, 0)$, semi-axes $4$ and $3$.

Q2. Find the asymptotes of $\frac{x^2}{4} - \frac{y^2}{1} = 1$. [2 marks]

• Cue. $y = \pm\frac{1}{2}x$.

Q3. Eliminate the parameter from $x = 1 + 2\cos t$, $y = 3 + 2\sin t$. [3 marks]

• Cue. $(x - 1)^2 + (y - 3)^2 = 4$, a circle of radius $2$ centred at $(1, 3)$.