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.

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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

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

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 \(c^2>2a^2\big).\)

\[x^2 + y^2 = 4l^2 \] \[x^2 + 4y^2 = l^2 \] \[4x^2 + y^2 = l^2 \] \[4x^2 + 4y^2 = l^2 \]

A rod of length \(l\) slides with its ends on the \(x\)-axis and \(y\)-axis.

Find the locus of its midpoint.

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

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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

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

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.

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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

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

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\().\)