Graphs of conics and parametric curves: VCE Specialist Mathematics Unit 3

Syllabus dot point

“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”

6 min readUpdated 29 May 2026

Quick answer

The ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ is centred at $(h, k)$ with semi-axes $a$ (horizontal) and $b$ (vertical). The hyperbola $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ opens left and right with vertices at $(h \pm a, k)$ and asymptotes $y - k = \pm\frac{b}{a}(x - h)$ . A parametric curve gives $x$ and $y$ each as a function of a parameter $t$ ; eliminating $t$ recovers the Cartesian equation. The standard parametrisation $x = a\cos t$ , $y = b\sin t$ traces the ellipse because $\cos^2 t + \sin^2 t = 1$ .

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What this dot point is asking
Ellipses
Hyperbolas
Parametric curves
Examples in context
Try this
Exam-style questions

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

\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.

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

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1.

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

y - k = \pm\frac{b}{a}(x - h).

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:

x = f(t), \qquad y = g(t).

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 $\sec^2 t - \tan^2 t = 1$ .

Worked example

From parametric to Cartesian

A curve is given by $x = 2 + 3\cos t$ and $y = -1 + 2\sin t$ , for $0 \le t < 2\pi$ . Find its Cartesian equation and describe the curve.

Step 1: Isolate the trigonometric expressions

Rearranging each equation to make $\cos t$ and $\sin t$ the subjects prepares them for substitution into the Pythagorean identity, which will eliminate the parameter $t$ .

\cos t = \frac{x - 2}{3}, \qquad \sin t = \frac{y + 1}{2}.

Step 2: Eliminate the parameter using the Pythagorean identity

The identity $\cos^2 t + \sin^2 t = 1$ holds for all $t$ , so substituting the expressions from Step 1 removes $t$ entirely and yields the Cartesian equation directly.

\left(\frac{x - 2}{3}\right)^2 + \left(\frac{y + 1}{2}\right)^2 = 1,

\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1.

Step 3: Identify the curve and its key features

Reading off the centre and semi-axes from the standard form $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ completes the description. A quick check at $t = 0$ verifies the parametrisation traces the correct ellipse.

This is an ellipse centred at $(2, -1)$ with $a = 3$ (semi-axis along $x$ ) and $b = 2$ (semi-axis along $y$ ). Its vertices are $(2 \pm 3, -1) = (-1, -1)$ and $(5, -1)$ horizontally, and $(2, -1 \pm 2) = (2, 1)$ and $(2, -3)$ vertically. Since $a > b$ the major axis is horizontal. Checking $t = 0$ gives $(5, -1)$ , a right-hand vertex, consistent with the equation.

Common exam traps

Mixing up $a^2$ and $a$: The denominators in the standard form are the squares of the semi-axes. The semi-axis is $a = \sqrt{a^2}$ , so $\frac{x^2}{9} + \frac{y^2}{4} = 1$ has semi-axes $3$ and $2$ , not $9$ and $4$ .
Wrong asymptote gradient: For $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ the asymptote gradients are $\pm\frac{b}{a}$ , not $\pm\frac{a}{b}$ .
Losing the parameter range: A parametric curve may be only part of the full Cartesian curve. Always check the values $x$ and $y$ actually take.
Treating a hyperbola as a function: A hyperbola fails the vertical line test, so it is a relation, not a function.

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 $(1, 3)$ .

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style2 marks

A track from point D to point O follows an elliptical path given by x = 2cos(t) + 2, y = (e - 2)sin(t), where t is in [-pi/2, pi/2]. Find the Cartesian equation of the elliptical path.

Show worked answer →

Eliminate the parameter t using the identity cos^2(t) + sin^2(t) = 1.

From x = 2cos(t) + 2, make cos(t) the subject: cos(t) = (x - 2) / 2.

From y = (e - 2)sin(t), make sin(t) the subject: sin(t) = y / (e - 2).

Substitute into cos^2(t) + sin^2(t) = 1:
((x - 2) / 2)^2 + (y / (e - 2))^2 = 1.

So the Cartesian equation is (x - 2)^2 / 4 + y^2 / (e - 2)^2 = 1.

This is an ellipse with centre (2, 0), semi-axis 2 in the x-direction and semi-axis (e - 2) in the y-direction (note e - 2 is about 0.718).