Maybe the most natural example to illustrate how useful loci are is definition of a parabola. As Wikipedia writes it is [i]the set of points equidistant from a single point (the focus) and a line (the directrix)[/i]. This definition is for beginners, however, not easy to handle. There is at least one abstract step in-between, namely that to find the distance from a line we may need a perpendicular being drawn.[br][br]Thus, when focus [color=#0000ff]F[/color] and directrix d are given, constructing one point P of parabola p is as follows:[br][br][list=1][*]Choose an arbitrary point [color=#7d7dff]D[/color] of [color=#0000ff]d[/color].[/*][*]Construct a perpendicular line to [color=#0000ff]d[/color] on [color=#7d7dff]D[/color].[/*][*]Construct the bisector b of [color=#7d7dff]D[/color] and [color=#0000ff]F[/color].[/*][*]Let the intersection point of the perpendicular line and b be P.[/*][*]Now P is a point of parabola [color=#ff0000]p[/color] since P[color=#7d7dff]D[/color]=P[color=#0000ff]F[/color] (because bisector b is actually the axis for the mirroring of point [color=#7d7dff]D[/color] to [color=#0000ff]F[/color]).[/*][/list][br]In fact drawing bisector b is also a hidden step since we implicitely used some basic properties of the reflection.[br][br]This kind of definition of the parabola is usual in many secondary schools, however the equivalence of this definition and the analytical one (that is, the usual formula for a parabola is [math]y=ax^2+bx+c[/math] for some constants a, b and c) is not obvious.

In the figure we can see a dark red dashed parabola which is the real locus drawn by GeoGebra numerically. The lighter red curve has been computed symbolically by the [b]LocusEquation[/b] command. In this special case these two curves are exactly the same. We will see some examples below where this is not the case. The reason comes from the algorithm we use to compute the equation. (In many cases the symbolical result can be improved by using extended algorithms.)

A parabola can also be obtained by getting the orthocenter points of a triangle if two vertices are fixed and the third one moves on a line which is parallel to the opposite side.

In this figure point [color=#7d7dff]A[/color] is constrained by line [color=#0000ff]PQ[/color]. Students can drag point [color=#7d7dff]A[/color] on line [color=#0000ff]PQ[/color] and see how the orthocenter [color=red]D[/color] is changing meanwhile. It is not obvious to prove that the locus here is a parabola, but the students are able at least to get experience by changing points [color=blue]B[/color] and [color=blue]C[/color] by preserving parallelism of lines [color=#0000ff]PQ[/color] and [color=#0000ff]BC[/color].[br][br]On the other hand, the locus equation will not be essentially different on other positions of [color=blue]P[/color], [color=blue]Q[/color], [color=blue]B[/color] and [color=blue]C[/color]: it will be quadratic in most set of positions. Some of these positions seem easy to investigate, for example when [color=#0000ff]PQ[/color] is perpendicular to [color=#0000ff]BC[/color] (here the result will be a linear equation). Others, for example by putting [color=blue]P[/color] to (1,1) and not changing anything else in the set, the locus result is a hyperbola, namely [math]x^2+xy+y=5[/math]. First, this formula is hard to analyze in secondary school since it is not in explicit form like a function [math]y=f(x)[/math]. Second, this formula is still a quadratic implicit equation and thus it can open horizons of generalization to cover all kind of conics.[br][br]In case [color=blue]P[/color]=(1,1), [color=blue]Q[/color]=(2,0), [color=blue]B[/color]=(3,1), [color=blue]C[/color]=(3,1) the computed locus equation is [math]-xy-x+y^2+3y=-2[/math], i.e. [math]-xy-x+y^2+3y+2=0[/math] whose left hand side is the product of [math](y+1)[/math] and [math](y-x+2)[/math], two lines, namely [math]y=-1[/math] and [math]y=x-2[/math] written in the usual explicit form. In this constellation height of side [color=#7d7dff]A[/color][color=blue]B[/color] always lies on line [color=#0000ff]CQ[/color] since it is perpendicular to [color=#0000ff]PQ[/color]. Thus point [color=red]D[/color] will also lie on line [color=#0000ff]CQ[/color], so it seems sensible that the locus equation is [color=#0000ff]CQ[/color] in this case. Unfortunately, GeoGebra's [b]LocusEquation[/b] shows an extra line here, not only [color=#0000ff]CQ[/color] (which has the explicit equation [math]y=x-2[/math]) but also another one. This example shows that the real locus may be a subset of the result of the [b]LocusEquation[/b] command.[br][br]One further step forward is to constrain point [color=#7d7dff]A[/color] on a circle, not a line. In this way we can obtain non-quadratic locus equations like the strophoid formula which is a cubic one.

Here points [color=#0000ff]P[/color] and [color=#0000ff]Q[/color] define a circle which has perimeter point [color=#7d7dff]A[/color] as a constraint.[br][br]In this figure a quartic equation is shown, but the real locus is a cubic curve. That is an extra component is shown (here line x=-1) as in many other cases when dragging points [color=#0000ff]P[/color], [color=#0000ff]Q[/color], [color=#0000ff]B[/color] or [color=#0000ff]C[/color]. On the other hand, by moving point [color=#0000ff]P[/color] down, for example into (-1,1) or (-1,0), the locus is the same as [b]LocusEquation[/b] computes: in these cases the locus is a real quartic curve.[br][br]A beautiful side case is when points when [color=blue]B[/color]=[color=blue]P[/color] and [color=blue]C[/color]=[color=blue]Q[/color]. In this case the real locus is a [url=http://mathworld.wolfram.com/RightStrophoid.html]right strophoid[/url] curve and an extra line component is drawn on points [color=blue]C[/color] and [color=blue]Q[/color].