Finding a locus: eliminate parameters and prove every point occurs

A locus is a set of attainable points

A locus describes all positions satisfying a stated construction or condition. Many elementary examples are curves; some problems ask for a region instead. In either case, the essential task is to prove both directions: every constructed point lies in your proposed set, and every point in that set is actually obtainable.

Two useful approaches are eliminating parameters and using geometry. Neither method automatically guarantees that an equation has no extra points.

Elimination must preserve restrictions

Suppose

$$x=\frac12t-1,\qquad y=t^2-3t+1,\qquad-1\le t\le1.$$

Since $t=2x+2$, substitution into both the equation and the interval yields

$$y=4x^2+2x-1,\qquad-\frac32\le x\le-\frac12.$$

Every $x$ in that interval corresponds to an allowed $t$, so this is the entire locus. Dropping the interval would incorrectly add the rest of the parabola. For more methods, see parametric curves.

Geometry can identify the curve, but check missing points

Drop a perpendicular from $A=(2,0)$ to the line $y=mx$, with $m$ ranging over all real numbers. Let the foot be $H$.

The right-angle condition places $H$ on the circle with diameter $OA$. However, the line never becomes vertical, so $H$ never reaches the origin. The position $H=A$ does occur when $m=0$, even though the corresponding right triangle is degenerate.

Algebra confirms the exact set. By orthogonal projection,

$$H=\left(\frac2{1+m^2},\frac{2m}{1+m^2}\right).$$

Thus

$$\boxed{(x-1)^2+y^2=1,\qquad(x,y)\ne(0,0).}$$

Conversely, a point on this circle other than the origin has $x>0$. Taking $m=y/x$ recovers its coordinates in the displayed projection formula. This proves that no other point is missing.

Example 1: transform a unit circle

Let $P=(x,y)$ move on $x^2+y^2=1$. Find the locus of

$$Q=(x+2y+1,\ 2x-y-1).$$

Write $Q=(X,Y)$ and solve for the original coordinates:

$$x=\frac{X+2Y+1}{5},\qquad y=\frac{2X-Y-3}{5}.$$

Substitution into the unit-circle equation gives

$$\frac{(X+2Y+1)^2+(2X-Y-3)^2}{25}=1,$$

or

$$\boxed{(X-1)^2+(Y+1)^2=5.}$$

The coordinate transformation is invertible, and each point on this new circle maps back to a point of the unit circle. Therefore the whole circle is attained.

Example 2: reflect a fixed point in a varying line

Let $A=(2,0)$ and let $B$ be its reflection in $y=mx$. Find the locus of $B$ as $m$ ranges over all real numbers. The source's prose omits the coordinates of $A$, but its diagram and solution use this fixed point.

The midpoint $M$ of $AB$ is the perpendicular foot just studied. Hence $M$ runs over the circle with diameter $OA$, excluding $O$. If $B=(X,Y)$, then $M=((X+2)/2,Y/2)$, giving

$$\left(\frac{X+2}{2}-1\right)^2+\left(\frac Y2\right)^2=1,$$

and the exclusion $M\ne O$ becomes $B\ne(-2,0)$. Thus

$$\boxed{X^2+Y^2=4,\qquad(X,Y)\ne(-2,0).}$$

The affine correspondence $B=2M-A$ is one-to-one, so every remaining point is attained. In particular $B=A$ occurs for the horizontal mirror line.

Example 3: midpoints of perpendicular radius vectors

Let $O$ be the origin and let $P,Q$ lie on $y=\sqrt3x^2$, with $\angle POQ=\pi/2$. Find the locus of the midpoint $R$ of $PQ$.

Set

$$P=(s,\sqrt3s^2),\qquad Q=(t,\sqrt3t^2),\qquad s,t\ne0.$$

The nonzero restriction is necessary because an angle with a zero vector is undefined. Orthogonality gives

$$\overrightarrow{OP}\cdot\overrightarrow{OQ}=st+3s^2t^2=0.$$

Since $st\ne0$, we obtain $st=-1/3$. Put $u=s+t$ and $v=st=-1/3$. For $R=(X,Y)$,

$$X=\frac u2,\qquad Y=\frac{\sqrt3}{2}(u^2-2v) =2\sqrt3X^2+\frac{\sqrt3}{3}.$$

To show that every real $X$ occurs, the required $s,t$ must be real, nonzero, and distinct. They are the roots of

$$z^2-2Xz-\frac13=0.$$

The discriminant is $4X^2+4/3>0$, and the product is $-1/3\ne0$. Thus these roots always meet all requirements and produce perpendicular nonzero vectors. Consequently the entire parabola

$$\boxed{Y=2\sqrt3X^2+\frac{\sqrt3}{3}}$$

is the locus.

The recurring checks are the original parameter interval, real-root conditions after symmetric substitutions, excluded degenerate constructions, and a converse showing that no proposed point is unattainable.