If R is any point on the x-axis and Q is any point on the y-axis, and P is a variable point on RQ with RP = b,

Exercise 6.1 · Q11

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Concept understanding — Locus of a Point

A point is not a thing but a place — a position located, with reference to a coordinate system, by a unique ordered pair $(x,y)$ : $x$ is the (signed) distance from the $y$ -axis and $y$ the (signed) distance from the $x$ -axis. The locus of a moving point is the path traced out by it as it moves subject to a stated geometric condition; equivalently, an equation in $x,y$ has infinitely many real solution-pairs $(x,y)$ , and the collection of all the points corresponding to those solutions is the locus of the equation. (The plural of locus is loci.)

Some standard loci: a point moving so as to stay equidistant from two fixed points $A,B$ traces the perpendicular bisector of $AB$ ; a point equidistant from two fixed lines traces the angle bisector; a point at a fixed distance $r$ from a fixed point $O$ traces a circle of radius $r$ centred at $O$ . A point on the rim of a circle rolling along a straight line traces a cycloid.

Procedure for finding the equation of a locus.

Assign coordinates $(h,k)$ to the moving point $P$ whose locus is wanted.
Translate the given geometric condition(s) into equation(s) relating $h,k$ and any parameters/known quantities.
Eliminate the parameter(s), so that only $h,k$ and known constants remain.
Replace $h$ by $x$ and $k$ by $y$ ; the resulting equation in $x,y$ is the equation of the locus.

Tip

When the given coordinates involve a trigonometric parameter $\theta$ (e.g. $(a\sec\theta,\,b\tan\theta)$ ), eliminate $\theta$ using a Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta-\tan^2\theta=1$ , or $\csc^2\theta-\cot^2\theta=1$ . When the parameter is a real number $t$ (e.g. $(ct,\,c/t)$ ), eliminate it by taking the product, ratio, or another algebraic combination of the two coordinate equations that cancels $t$ . …

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Use the section formula to locate $P$ between $R$ and $Q$ , then eliminate $r,q$ using the fact that $RQ=RP+PQ=a+b$ .

Step 1. Set up $R,Q$ and the ratio. Let $R=(r,0)$ on the $x$ -axis and $Q=(0,q)$ on the $y$ -axis. $P$ lies on segment $RQ$ with $RP=b$ and $PQ=a$ , so $P$ divides $RQ$ (measuring from $R$ to $Q$ ) in the ratio

RP:PQ = b:a.

Step 2. Apply the section formula. For $P=(x,y)$ dividing $R(r,0)$ to $Q(0,q)$ internally in ratio $b:a$ (from $R$ ):

x=\frac{b\cdot 0+a\cdot r}{a+b}=\frac{ar}{a+b}, \qquad y=\frac{b\cdot q+a\cdot 0}{a+b}=\frac{bq}{a+b}.

Step 3. Solve for $r,q$ .

r=\frac{x(a+b)}{a}, \qquad q=\frac{y(a+b)}{b}.

Step 4. Use the fixed total length $RQ=a+b$ . Since $R=(r,0)$ and $Q=(0,q)$ , …

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Q.If RRR is any point on the xxx-axis and QQQ is any point on the yyy-axis, and PPP is a variable point on RQRQRQ with RP=bRP = bRP=b, PQ=aPQ = aPQ=a, then find the equation of the locus of PPP.

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Q.If $R$ is any point on the $x$ -axis and $Q$ is any point on the $y$ -axis, and $P$ is a variable point on $RQ$ with $RP = b$ , $PQ = a$ , then find the equation of the locus of $P$ .

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