Question 303 marksShort AnswerSection C

  1. (A) If x=h+acos⁡θx = h + a \cos \theta, y=k+bsin⁡θy = k + b \sin \theta, then prove that : (x−ha)2+(y−kb)2=1\left(\dfrac{x - h}{a}\right)^2 + \left(\dfrac{y - k}{b}\right)^2 = 1.

Introduction to Trigonometry
Eliminating a parameter using an identity

Official answer

Proved: from the given equations, (x−h)/a=cos⁡θ(x - h)/a = \cos \theta and (y−k)/b=sin⁡θ(y - k)/b = \sin \theta. Squaring both and adding gives (x−ha)2+(y−kb)2=cos⁡2θ+sin⁡2θ=1\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta = 1 (by the fundamental trigonometric identity), which is the required relation. Geometrically, this is the standard Cartesian equation of the ellipse whose parametric form is given, centred at (h,k)(h, k) with semi-axes a and b.

Answer keywords:parameter eliminationcos θ = (x−h)/asin θ = (y−k)/bsquare and addsin²θ + cos²θ = 1Pythagorean identityequation of ellipseprove that

Marking scheme

  1. 1.1 mark: rearranging the given equations to (x−h)/a=cos⁡θ(x - h)/a = \cos \theta and (y−k)/b=sin⁡θ(y - k)/b = \sin \theta.
  2. 2.1 mark: squaring both expressions correctly to get (x−ha)2=cos⁡2θ\left(\dfrac{x - h}{a}\right)^2 = \cos^2\theta and (y−kb)2=sin⁡2θ\left(\dfrac{y - k}{b}\right)^2 = \sin^2\theta.
  3. 3.1 mark: adding and applying sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to conclude the left side equals 1 (hence proved).

Hint

Make cos θ and sin θ the subjects: (x−h)/a=cos⁡θ(x-h)/a = \cos \theta and (y−k)/b=sin⁡θ(y-k)/b = \sin \theta; square both, add, and use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

Quick oral answer

I rewrite the equations as cos⁡θ=(x−h)/a\cos \theta = (x-h)/a and sin⁡θ=(y−k)/b\sin \theta = (y-k)/b, square both and add; the right side becomes cos⁡2θ+sin⁡2θ\cos^2\theta + \sin^2\theta which equals 11, so (x−ha)2+(y−kb)2=1\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = 1 is proved.

Common mistakes

  1. 1.Squaring the equations before dividing by a and b, leaving a2a^2 and b2b^2 tangled with the trig terms instead of cleanly getting cos⁡2θ\cos^2\theta and sin⁡2θ\sin^2\theta.
  2. 2.Squaring only one of the two equations, or adding without squaring, so the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 cannot be applied.
  3. 3.Writing sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta as something other than 1 (e.g. leaving it unsimplified) and thus not completing the proof.

Analysis and explanation

Eliminate the parameter θ between the two given equations using the Pythagorean identity to reach the required relation.


Concept

  • Isolate cos θ and sin θ: (x−h)/a=cos⁡θ(x - h)/a = \cos \theta and (y−k)/b=sin⁡θ(y - k)/b = \sin \theta.
  • Square both expressions and add; the right side becomes cos⁡2θ+sin⁡2θ\cos^2\theta + \sin^2\theta, which the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 collapses to 11.
  • This proves (x−ha)2+(y−kb)2=1\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = 1.

Common mistakes

  • Squaring before dividing by a and b (must isolate cos θ/sin θ first).
  • Squaring only one of the two equations, or treating a and b as trigonometric quantities instead of constants moved to the denominator.

Real-world

  • x=h+acos⁡θx = h + a \cos \theta, y=k+bsin⁡θy = k + b \sin \theta are the parametric equations of an ellipse centred at (h,k)(h, k) with semi-axes a and b; the proved relation is exactly its standard Cartesian equation — the same algebra used to describe planetary orbits.

Interesting facts

The relation proved, (x−ha)2+(y−kb)2=1\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = 1, is the standard equation of an ellipse centred at (h,k)(h, k) with semi-axes a and b — so this problem is secretly deriving the ellipse from its parametric form.

Parametric equations like x=h+acos⁡θx = h + a \cos \theta, y=k+bsin⁡θy = k + b \sin \theta are how computers and animators trace out smooth curves, and how astronomers describe elliptical planetary orbits following Kepler's first law.

Question details

Q30 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Introduction to Trigonometry
Topic
Eliminating a parameter using an identity
Bloom's level
apply
Syllabus topic
Unit 5: Trigonometry — Introduction to Trigonometry: Proving relations by eliminating θ using sin²θ + cos²θ = 1 (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why must I divide by a and b before squaring?

To isolate the pure trig ratios cos θ and sin θ. Only then does squaring give cos⁡2θ\cos^2\theta and sin⁡2θ\sin^2\theta, which the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 can simplify. Squaring first would leave a2a^2 and b2b^2 attached and block the identity.

Which identity is used to eliminate θ?

The fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Once cos θ and sin θ are squared and added, the right-hand side becomes exactly this and equals 1.

What does the final equation represent?

It is the standard Cartesian equation of an ellipse with centre (h, k) and semi-axes a and b, so the problem effectively converts the parametric form of an ellipse into its Cartesian form.