Q30
3 marksShort AnswerSection C

If x=h+acos⁡θ,y=k+bsin⁡θx = h + a\cos\theta, y = k + b\sin\theta, then prove that :

(x−ha)2+(y−kb)2=1\left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = 1

Introduction to Trigonometry
Trigonometric Identities (Parametric Form of an Ellipse)
Official Answer

Proved: (x−ha)2+(y−kb)2=1\left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = 1. From x=h+acos⁡θx = h + a\cos\theta, x−ha=cos⁡θ\frac{x-h}{a} = \cos\theta; from y=k+bsin⁡θy = k + b\sin\theta, y−kb=sin⁡θ\frac{y-k}{b} = \sin\theta. Squaring both, (x−ha)2=cos⁡2θ\left(\frac{x-h}{a}\right)^2 = \cos^2\theta and (y−kb)2=sin⁡2θ\left(\frac{y-k}{b}\right)^2 = \sin^2\theta. Adding, (x−ha)2+(y−kb)2=cos⁡2θ+sin⁡2θ\left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta. Since sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 for all θ, the sum equals 1, proving the required identity holds for every value of θ.

parametric equationstrigonometric identitysin squared plus cos squared equals oneeliminate parameter thetaellipse parametric formalgebraic proof

Marking Scheme

  • 11 mark: correctly isolating x−ha=cos⁡θ\frac{x-h}{a} = \cos\theta and y−kb=sin⁡θ\frac{y-k}{b} = \sin\theta from the given equations.
  • 21 mark: correctly squaring both expressions to get (x−ha)2=cos⁡2θ\left(\frac{x-h}{a}\right)^2 = \cos^2\theta and (y−kb)2=sin⁡2θ\left(\frac{y-k}{b}\right)^2 = \sin^2\theta.
  • 31 mark: correctly adding the two squared expressions and applying the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to reach the required result.

Hint

Isolate cos θ from the x-equation and sin θ from the y-equation, square both, and add using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

Quick Oral Answer

I isolate cos θ and sin θ from the two given equations, square each, and add them; since sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 always, the sum equals 1, which proves the required identity.

Analysis & Explanation

Eliminate θ from the two parametric equations using the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.


Concept

  • Isolate cos θ and sin θ from the given equations, then square and add.
  • These equations are the parametric form of an ellipse centred at (h, k) with semi-axes a, b.

Key Points

  • x−ha=cos⁡θ\frac{x-h}{a} = \cos\theta and y−kb=sin⁡θ\frac{y-k}{b} = \sin\theta; squaring and adding gives cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1.

Common Mistakes

  • Substituting a specific numeric value of θ instead of proving the identity holds for all θ.
  • Sign/division errors while isolating cos θ or sin θ.

Real-world Application

  • This parametrisation links trigonometry to conics (circle when a=ba = b), used in describing elliptical orbits and elliptical arches/gears in engineering.

Common Mistakes

  1. 1Substituting a specific value of θ (like 0° or 90°) to 'verify' rather than proving the identity algebraically for a general θ.
  2. 2Forgetting to divide by a and b respectively while isolating cos θ and sin θ, leading to an incorrect squared expression.
  3. 3Sign or arrangement errors when subtracting h and k from x and y respectively before isolating the trigonometric ratios.

Interesting Facts

The given parametric equations x=h+acos⁡θ,y=k+bsin⁡θx = h + a\cos\theta, y = k + b\sin\theta actually represent an ellipse centred at (h, k) with semi-major/minor axes a and b — when a=ba = b, this reduces to the parametric form of a circle.

Parametric equations, using a third variable (parameter) like θ to define x and y, are widely used in physics and engineering to describe motion along curved paths, such as projectile motion or planetary orbits.

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 used here is one of the three fundamental trigonometric identities and is essentially a restatement of the Pythagoras theorem for a right triangle inscribed in a unit circle.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

What geometric curve do these parametric equations represent?

They represent an ellipse centred at (h, k) with semi-axis lengths a (along x) and b (along y); if a = b, the curve becomes a circle of radius a centred at (h, k).

Can this proof be done by substituting a particular angle for θ?

No — since the question says 'prove that', the identity must hold for every value of θ, so it must be shown algebraically using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, not verified for one specific angle only.