Locus, parabolas, circles, ellipses and hyperbolas (Cartesian, parametric, polar)
Master locus, parabolas, circles, ellipses and hyperbolas in Year 11 VCE Specialist Mathematics. A locus is the set of all points that satisfy a geometric condition, and the four conic sections — the circle \(x^2+y^2=r^2\), the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), the parabola \(y^2=4ax\) and the hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) — each have Cartesian, parametric and polar forms. It sits in the Functions, relations and graphs area of study of the VCE Mathematics Study Design (VCAA), within the Functions, relations and graphs topic of Unit 2.
You will learn to obtain an equation from a locus definition using the distance formula, read off key features such as centre, radius, semi-axes and vertices and the linear asymptotes \(y=\pm\dfrac{b}{a}x\) of a hyperbola, convert between parametric and Cartesian form, and change between polar and Cartesian coordinates with \(x=r\cos\theta,\ y=r\sin\theta\) — the groundwork for curve sketching and coordinate geometry across the course.
Theory
A locus is the set of all points that satisfy a given geometric condition. In Year 11 Specialist Mathematics you describe circles, ellipses, parabolas and hyperbolas by their Cartesian equations, parametric forms and polar forms; read off key features such as the centre, radius, semi-axes, vertices and asymptotes; and use a distance or locus definition to obtain an equation. This page shows how, with fully worked examples.
A locus is the path traced by all points in the plane that meet a stated geometric condition. Turning that condition into an equation with the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) is how the curves below are built. The four conics — circle, ellipse, parabola and hyperbola — each have a Cartesian, a parametric and a polar description.
A circle is the locus of points a fixed distance \(r\) (the radius) from a fixed point \((h,k)\) (the centre): \((x-h)^2+(y-k)^2=r^2\). A convenient parametric form is \(x=h+r\cos t,\ y=k+r\sin t\), since \(\cos^2 t+\sin^2 t=1\) recovers the Cartesian equation.
An ellipse centred at the origin has equation \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\). It reaches \(x=\pm a\) and \(y=\pm b\), so \(a\) and \(b\) are the semi-axes; the larger is the semi-major axis. Its parametric form is \(x=a\cos t,\ y=b\sin t\).
A parabola in the form \(y^2=4ax\) has its vertex at the origin and opens along the \(x\)-axis, with parametric form \(x=at^2,\ y=2at\). (A parabola may also be written \(y=ax^2\), opening along the \(y\)-axis.)
A hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) has two branches with vertices at \((\pm a,0)\) and two linear asymptotes \(y=\pm\dfrac{b}{a}x\) that the branches approach far from the centre.
Polar coordinates \((r,\theta)\) give a point by its distance \(r\) from the origin and the angle \(\theta\) from the positive \(x\)-axis. They convert to and from Cartesian coordinates by \(x=r\cos\theta,\ y=r\sin\theta\) and \(r^2=x^2+y^2,\ \tan\theta=\dfrac{y}{x}\). For example the polar equation \(r=k\) is a circle of radius \(k\) centred at the origin.
The circle with centre \((h,k)\) and radius \(r\), in Cartesian then parametric form:
The ellipse centred at the origin with semi-axes \(a\) and \(b\):
The parabola with vertex at the origin, opening along the \(x\)-axis:
The hyperbola centred at the origin, with its two linear asymptotes and vertices at \((\pm a,0)\):
The polar coordinates \((r,\theta)\) convert to and from Cartesian coordinates by:
Working with loci and conics
- From a locus / distance definition: write the geometric condition as an equation using the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), then simplify. "A fixed distance \(r\) from \((h,k)\)" gives \((x-h)^2+(y-k)^2=r^2\).
- Identify the conic and read its features: match the equation to a standard form. Complete the square if needed to find a circle's centre and radius; take square roots of the denominators for an ellipse's semi-axes or a hyperbola's \(a,b\); read \(4a\) from a parabola.
- Parametric \(\to\) Cartesian: eliminate the parameter \(t\). For circles and ellipses use \(\cos^2 t+\sin^2 t=1\); for a parabola make \(t\) the subject of one equation and substitute into the other.
- Cartesian \(\to\) parametric: read \(a\) and \(b\) from the standard form and write \(x=a\cos t,\ y=b\sin t\) (ellipse/circle) or \(x=at^2,\ y=2at\) (parabola).
- Polar \(\leftrightarrow\) Cartesian: to go to Cartesian use \(x=r\cos\theta,\ y=r\sin\theta\); to go to polar use \(r=\sqrt{x^2+y^2}\) and \(\tan\theta=\dfrac{y}{x}\), choosing \(\theta\) in the correct quadrant.
For a hyperbola, always state the linear asymptotes \(y=\pm\dfrac{b}{a}x\) as part of describing its graph — they are a required key feature.
The fixed distance is the radius: the distance from the centre \((1,2)\) to the point \((4,6)\). Use \(r^2=(x_2-x_1)^2+(y_2-y_1)^2\):
| \(r^2\) | \(=\) | \((4-1)^2+(6-2)^2\) |
| \(=\) | \(3^2+4^2\) | |
| \(=\) | \(9+16\) | |
| \(=\) | \(25\) |
Write the locus in the form \((x-h)^2+(y-k)^2=r^2\) with centre \((1,2)\):
| \((x-1)^2+(y-2)^2\) | \(=\) | \(25\) |
The locus is the circle \((x-1)^2+(y-2)^2=25\).
Group the \(x\)- and \(y\)-terms and complete the square:
| \(x^2+y^2-6x+4y-12\) | \(=\) | \(0\) |
| \((x^2-6x)+(y^2+4y)\) | \(=\) | \(12\) |
| \((x-3)^2-9+(y+2)^2-4\) | \(=\) | \(12\) |
| \((x-3)^2+(y+2)^2\) | \(=\) | \(25\) |
Read off the centre \((h,k)\) and radius \(r=\sqrt{r^2}\):
| \((h,k)\) | \(=\) | \((3,-2)\) |
| \(r\) | \(=\) | \(\sqrt{25}=5\) |
Test the point \((7,1)\) in the left-hand side:
| \((7-3)^2+(1+2)^2\) | \(=\) | \(4^2+3^2\) |
| \(=\) | \(16+9\) | |
| \(=\) | \(25\ \checkmark\) |
Centre \((3,-2)\), radius \(5\); and \((7,1)\) lies on the circle.
Make \(\cos t\) and \(\sin t\) the subjects:
| \(\cos t\) | \(=\) | \(\dfrac{x}{2}\) |
| \(\sin t\) | \(=\) | \(\dfrac{y}{5}\) |
Substitute into \(\cos^2 t+\sin^2 t=1\) to eliminate \(t\):
| \(\cos^2 t+\sin^2 t\) | \(=\) | \(1\) |
| \(\dfrac{x^2}{4}+\dfrac{y^2}{25}\) | \(=\) | \(1\) |
The curve is the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{25}=1\), with semi-axes \(2\) (along \(x\)) and \(5\) (along \(y\)); the semi-major axis is \(5\).
Make \(t\) the subject of \(y=2t\), then substitute into \(x=t^2\):
| \(t\) | \(=\) | \(\dfrac{y}{2}\) |
| \(x\) | \(=\) | \(\left(\dfrac{y}{2}\right)^2=\dfrac{y^2}{4}\) |
| \(y^2\) | \(=\) | \(4x\) |
The Cartesian equation is \(y^2=4x\), a parabola with vertex at the origin opening to the right.
Identify \(a\) and \(b\) from \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\):
| \(a^2\) | \(=\) | \(4\Rightarrow a=2\) |
| \(b^2\) | \(=\) | \(9\Rightarrow b=3\) |
The vertices lie on the \(x\)-axis at \(x=\pm a\):
| \((\pm a,\,0)\) | \(=\) | \((\pm 2,\,0)\) |
The asymptotes are \(y=\pm\dfrac{b}{a}x\):
| \(y\) | \(=\) | \(\pm\dfrac{b}{a}\,x\) |
| \(=\) | \(\pm\dfrac{3}{2}\,x\) |
\(a=2\); vertices \((\pm2,0)\); asymptotes \(y=\pm\dfrac{3}{2}x\).
Find \(r\) using \(r=\sqrt{x^2+y^2}\):
| \(r\) | \(=\) | \(\sqrt{1^2+(\sqrt3)^2}\) |
| \(=\) | \(\sqrt{1+3}\) | |
| \(=\) | \(\sqrt4=2\) |
Find \(\theta\) from \(\tan\theta=\dfrac{y}{x}\) (the point is in the first quadrant):
| \(\tan\theta\) | \(=\) | \(\dfrac{\sqrt3}{1}=\sqrt3\) |
| \(\theta\) | \(=\) | \(\dfrac{\pi}{3}\) |
The polar coordinates are \(\left(2,\ \dfrac{\pi}{3}\right)\).
Common pitfalls
Frequently asked questions
What is a locus?
A locus is the set of all points in the plane that satisfy a given geometric condition. For example, the set of points a fixed distance \(r\) from a fixed point is a circle of radius \(r\).
What are the standard equations of the conics?
A circle is \((x-h)^2+(y-k)^2=r^2\); an ellipse is \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\); a parabola is \(y^2=4ax\) (or \(y=ax^2\)); and a hyperbola is \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\).
How do I find a circle’s centre and radius from its general equation?
Complete the square in \(x\) and in \(y\) to reach \((x-h)^2+(y-k)^2=r^2\). Then the centre is \((h,k)\) and the radius is \(r=\sqrt{r^2}\).
How do I convert a parametric form to a Cartesian equation?
Eliminate the parameter \(t\). For a circle or ellipse, make \(\cos t\) and \(\sin t\) the subjects and use \(\cos^2 t+\sin^2 t=1\). For a parabola such as \(x=t^2,\ y=2t\), make \(t\) the subject of one equation and substitute into the other.
What are the asymptotes of a hyperbola?
For \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) the two linear asymptotes are \(y=\pm\dfrac{b}{a}x\). The branches approach these straight lines far from the centre, so they must be stated as a key feature of the graph.
How do polar and Cartesian coordinates convert?
From polar to Cartesian, \(x=r\cos\theta\) and \(y=r\sin\theta\). From Cartesian to polar, \(r=\sqrt{x^2+y^2}\) and \(\tan\theta=\dfrac{y}{x}\), with \(\theta\) chosen in the correct quadrant.