(A) If $x = h + a \cos \theta$ , $y = k + b \sin \theta$ , then prove that : $\left(\dfrac{x - h}{a}\right)^2 + \left(\dfrac{y - k}{b}\right)^2 = 1$ .

Introduction to Trigonometry

Eliminating a parameter using an identity

Official Answer

Proved: from the given equations, $(x - h)/a = \cos \theta$ and $(y - k)/b = \sin \theta$ . Squaring both and adding gives $\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta = 1$ (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 $(h, k)$ with semi-axes a and b.

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

Marking Scheme

11 mark: rearranging the given equations to $(x - h)/a = \cos \theta$ and $(y - k)/b = \sin \theta$ .
21 mark: squaring both expressions correctly to get $\left(\dfrac{x - h}{a}\right)^2 = \cos^2\theta$ and $\left(\dfrac{y - k}{b}\right)^2 = \sin^2\theta$ .
31 mark: adding and applying $\sin^2\theta + \cos^2\theta = 1$ to conclude the left side equals 1 (hence proved).

Hint

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

Quick Oral Answer

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

Analysis & 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: $(x - h)/a = \cos \theta$ and $(y - k)/b = \sin \theta$ .
Square both expressions and add; the right side becomes $\cos^2\theta + \sin^2\theta$ , which the identity $\sin^2\theta + \cos^2\theta = 1$ collapses to $1$ .
This proves $\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = 1$ .

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

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

Common Mistakes

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

Interesting Facts

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

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

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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 $\cos^2\theta$ and $\sin^2\theta$ , which the identity $\sin^2\theta + \cos^2\theta = 1$ can simplify. Squaring first would leave $a^2$ and $b^2$ attached and block the identity.

Which identity is used to eliminate θ?

The fundamental Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ . 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.

(A) If x=h+acos⁡θx = h + a \cos \thetax=h+acosθ, y=k+bsin⁡θy = k + b \sin \thetay=k+bsinθ, 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(ax−h​)2+(by−k​)2=1.