Find the points on the locus of points that are 3 units from the x-axis and 5 units from the point (5, 1).

Exercise 6.1 · Q14

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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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Combine the two conditions — constant distance from the $x$ -axis and constant distance from a fixed point — and solve the resulting system for actual coordinates.

Step 1. Translate "3 units from the $x$ -axis". This gives $y=3$ or $y=-3$ .

Step 2. Translate "5 units from $(5,1)$ ". For a point $(x,y)$ ,

(x-5)^2+(y-1)^2=5^2=25.

Step 3. Case $y=3$ . Substitute $y=3$ :

(x-5)^2+(3-1)^2=25 \;\Rightarrow\; (x-5)^2+4=25 \;\Rightarrow\; (x-5)^2=21.

x-5=\pm\sqrt{21} \;\Rightarrow\; x=5\pm\sqrt{21}.

This gives the points $(5-\sqrt{21},\,3)$ and $(5+\sqrt{21},\,3)$ .

Step 4. Case $y=-3$ . Substitute $y=-3$ :

$(x-5)^2+(-3-1)^2=25 \;\Rightarrow\; (x-5)^2+16=25 \;\Rightarrow\; (x-5)^2=9.$ …

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Q.Find the points on the locus of points that are 333 units from the xxx-axis and 555 units from the point (5,1)(5, 1)(5,1).

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Q.Find the points on the locus of points that are $3$ units from the $x$ -axis and $5$ units from the point $(5, 1)$ .

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