Q.If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are x=acos3θ, y=asin3θ.
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.
When the given coordinates involve a trigonometric parameter θ (e.g. (asecθ,btanθ)), eliminate θ using a Pythagorean identity: sin2θ+cos2θ=1, sec2θ−tan2θ=1, or csc2θ−cot2θ=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.
Worked pattern. For P equidistant from two fixed points A(x1,y1) and B(x2,y2): set P=(h,k), write PA=PB, square both sides, and expand — the h2,k2 terms cancel, leaving a linear equation (the perpendicular bisector). For P at a fixed distance from a fixed point, PA=r squared gives the circle equation directly. For a parametric locus (f(θ),g(θ)), solve each coordinate equation for the trig function of θ it contains, then combine using an identity to eliminate θ entirely — this is exactly how the standard hyperbola/ellipse/astroid forms arise from their parametric descriptions.
The phrase "locus of a point class 11 maths definition and examples" shows up often in student searches tied to the Straight Lines chapter of the NCERT/CBSE Class 11 Mathematics curriculum, where locus problems are a recurring important-questions category in both board exams and JEE Main. Learning the standard four-step procedure here also pays off directly in later chapters on circles and conic sections that build on the same idea.
Write cosθ=(x/a)1/3, sinθ=(y/a)1/3 and eliminate θ via cos2θ+sin2θ=1, raised to the power 2/3 term-wise.
x2/3+y2/3=a2/3
Isolate cosθ and sinθ, then use cos2θ+sin2θ=1.
Step 1. Set up the coordinate equations. x=acos3θ and y=asin3θ, so
ax=cos3θ,ay=sin3θ.
Step 2. Solve for cosθ and sinθ. Taking the cube root of each,
cosθ=(ax)1/3,sinθ=(ay)1/3.
Step 3. Square both and add. cos2θ=(ax)2/3 and sin2θ=(ay)2/3, and since cos2θ+sin2θ=1:
(ax)2/3+(ay)2/3=1.
Step 4. Clear the common denominator. Multiplying through by a2/3:
x2/3+y2/3=a2/3.
This curve is the astroid.
x2/3+y2/3=a2/3
Take cube roots to isolate cosθ,sinθ, then apply the Pythagorean identity
- Squaring x=acos3θ directly instead of first taking the cube root, which does not eliminate θ
- Dropping the 2/3 power and writing x2+y2=a2 (confusing the astroid with a circle)
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following equations is the locus of (acosθ,bsinθ)?(a) x2+y2=a2(b) a2x2−b2y2=1(c) y2=4ax(d) a2x2+b2y2=1
›Reveal solutionSolution
With x=acosθ, y=bsinθ, using cos2θ+sin2θ=1 eliminates θ to give the ellipse a2x2+b2y2=1.
Let x=acosθ and y=bsinθ. Then cosθ=ax and sinθ=by.
Using the Pythagorean identity cos2θ+sin2θ=1:
(ax)2+(by)2=1⇒a2x2+b2y2=1
This is the standard equation of an ellipse — the point (acosθ,bsinθ) is the well-known parametric form of a point on an ellipse.
✓Final answerThe correct option is (d) a2x2+b2y2=1.
- CBSE 2024Set ANNUAL1 markMCQQ.If the point (8,−5) lies on the locus 16x2−25y2=k, then the value of k is:(a) 2(b) 0(c) 3(d) 1
›Reveal solutionSolution
Substituting (8,−5) into the locus equation gives k=3.
The locus is 16x2−25y2=k. Since (8,−5) lies on it, put x=8, y=−5:
1682−25(−5)2=1664−2525=4−1=3.
So k=3.
✓Final answerk=3 — option (c).
- CBSE 2023Set ANNUAL1 markMCQQ.The points lie on the locus of 3x2+3y2−8x−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=0. ✗
Test (0,−1): 0+3−0+12+17=32=0. ✗
Test (−2,3): 12+27+16−36+17=36=0. ✗
Only (1,2) lies on the locus.
✓Final answer(1,2).
- CBSE 2020Set ANNUAL1 markMCQQ.If the point (8,−5) lies on the locus 16x2−25y2=k, then the value of k is:(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:
16(8)2−25(−5)2=1664−2525=4−1=3.
So k=3.
✓Final answerThe correct option is (d) 3.
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