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

R

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

9% · 11/129 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $t$ . …

Start your 14-day free trial to unlock the full solution →

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 $Q=(0,q)$ , …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$(a\cos\theta, b\sin\theta)$
(a) $x^2+y^2=a^2$
(b) $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
(c) $y^2=4ax$
(d) $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
›Reveal solutionSolution

With $x=a\cos\theta$ , $y=b\sin\theta$ , using $\cos^2\theta+\sin^2\theta=1$ eliminates $\theta$ to give the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ .

Let $x=a\cos\theta$ and $y=b\sin\theta$ . Then $\cos\theta=\dfrac{x}{a}$ and $\sin\theta=\dfrac{y}{b}$ .

Using the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ :

$\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1 \;\Rightarrow\; \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$(8,-5)$
(a) $2$
(b) $0$
(c) $3$
(d) $1$
›Reveal solutionSolution

Substituting $(8,-5)$ into the locus equation gives $k=3$ .

The locus is $\dfrac{x^2}{16}-\dfrac{y^2}{25}=k$ . Since $(8,-5)$ lies on it, put $x=8,\ y=-5$ : …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$3x^2 + 3y^2 - 8x - 12y + 17 = 0$
(a) (1, 2)
(b) (0, 0)
(c) (0, -1)
(d) (-2, 3)
›Reveal solutionSolution

Substitute each option into the equation; the point that makes it exactly zero lies on the curve.

Test $(1,2)$ : $3(1)^2+3(2)^2-8(1)-12(2)+17 = 3+12-8-24+17 = 0$ . ✓ It satisfies the equation.

Test $(0,0)$ : $0+0-0-0+17=17\ne 0$ . ✗

…

Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 markMCQ
$(8, -5)$
(a) 0
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution

Direct substitution of the point $(8,-5)$ into the locus equation gives $k=3$ .

A point lies on a locus if its coordinates satisfy the locus equation. Substituting $x=8$ , $y=-5$ : …

Start your 14-day free trial to see the full solution →

←Previous question · 6.2Q10 — If P(2, -7) is a given point and Q is a point on 2x² + 9y² = 18, then …→Next question · 6.2Q12 — If the points P(6, 2) and Q(-2, 1), and R, are the vertices of a trian…

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.

🎓Unlock everything free for 14 days

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 $P$ .

Unlock everything free for 14 days