Equation of Locus

A locus is a set of points which satisfy certain geometric conditions. Many geometric shapes are most naturally and easily described as loci. For example, a circle is the set of points in a plane which are a fixed distance $r$ from a given point $P,$ the center of the circle.

Problems involving describing a certain locus can often be solved by explicitly finding equations for the coordinates of the points in the locus. Here is a step-by-step procedure for finding plane loci:

Step 1: If possible, choose a coordinate system that will make computations and equations as simple as possible.
Step 2: Write the given conditions in a mathematical form involving the coordinates $x$ and $y$.
Step 3: Simplify the resulting equations.
Step 4: Identify the shape cut out by the equations.

Step 1 is often the most important part of the process since an appropriate choice of coordinates can simplify the work in steps 2-4 immensely.

Find the locus of points $P$ such that the sum of the squares of the distances from $P$ to $A$ and from $P$ to $B,$ where $A$ and $B$ are two fixed points in the plane, is a fixed positive constant.

---

After rotating and translating the plane, we may assume that $A = (-a,0)$ and $B = (a,0).$ Suppose the constant is $c^2,$ $c\ne 0.$ Then

$$\begin{aligned} PA^2 + PB^2 &= c^2 \\ (x+a)^2+y^2+(x-a)^2+y^2 &= c^2 \\ 2x^2+2y^2+2a^2 &= c^2 \\ x^2+y^2 &= \frac{c^2}{2}-a^2. \end{aligned}$$

So the locus is either empty $\big($if $c^2 < 2a^2\big),$ a point $\big($if $c^2=2a^2\big),$ or a circle $\big($if $c^2>2a^2\big).$

Describe the locus of the points in a plane which are equidistant from a line and a fixed point not on the line.

---

After rotation and translation (and possibly reflection), we may assume that the point is $(0,2a)$ with $a\ne 0$ and that the line is the $x$-axis. The distance from $(x,y)$ to the $x$-axis is $|y|,$ and the distance to the point is $\sqrt{x^2 + (y-2a)^2},$ so the equation becomes

$$\begin{aligned} y^2 &= x^2+(y-2a)^2 \\ 0 &= x^2-4ay+4a^2 \\ y &= \frac{x^2}{4a} + a, \end{aligned}$$

which describes a parabola.

Note that if the point did lie on the line, e.g. $a=0,$ the equation reduces to $x^2=0,$ or $x=0,$ which gives a line perpendicular to the original line through the point; this makes sense geometrically as well. $_\square$

Find the locus of all points $P$ in a plane such that the sum of the distances $PA$ and $PB$ is a fixed constant, where $A$ and $B$ are two fixed points in the plane.

---

After translating and rotating, we may assume $A = (-a,0)$ and $B = (a,0),$ and let the constant be $c.$ If $c < 2a,$ then the locus is clearly empty, and if $c=2a,$ then the locus is a point, so assume $c>2a.$ Let $PA = d_1$ and $PB=d_2.$ Then $d_1^2+d_2^2 = (x+a)^2+y^2+(x-a)^2+y^2 = 2x^2+2y^2+2a^2,$ and $d_1^2-d_2^2 = 4ax.$ The locus equation is

$$\begin{aligned} d_1+d_2 &= c \\ d_1^2+d_2^2+2d_1d_2 &= c^2 \\ 4d_1^2d_2^2 &= \big(c^2-d_1^2-d_2^2\big)^2 \\ 4d_1^2d_2^2 &= c^4 - 2c^2\big(d_1^2+d_2^2\big) + \big(d_1^2+d_2^2\big)^2 \\ 0 &= c^4-2c^2\big(d_1^2+d_2^2\big) + \big(d_1^2-d_2^2\big)^2 \\ 0 &= c^4-2c^2\big(2x^2+2y^2+2a^2\big)+16a^2x^2 \\ \big(4c^2-16a^2\big)x^2+\big(4c^2\big)y^2 &= c^2\big(c^2-4a^2\big). \end{aligned}$$

Since $4c^2-16a^2>0$ and $c^2-4a^2>0,$ this is the equation of an ellipse. $_\square$

Note that if $a=0,$ this describes a circle, as expected $(A$ and $B$ coincide$).$