If $x = h + a\cos\theta$, $y = k + b\sin\theta$, then prove that : $\left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = 1$
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
Given: $x = h + a\cos\theta$, $y = k + b\sin\theta$
To Prove: $\left(\dfrac{x-h}{a}\right)^2 + \left(\dfrac{y-k}{b}\right)^2 = 1$
Proof:
From the given equations:
$$x - h = a\cos\theta \implies \frac{x-h}{a} = \cos\theta$$
$$y - k = b\sin\theta \implies \frac{y-k}{b} = \sin\theta$$
Now, consider the LHS:
$$\left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta$$
Using the identity $\sin^2\theta + \cos^2\theta = 1$:
$$= 1 = \text{ RHS}$$
Hence proved. $\blacksquare$
Source: Chapter 8, Section 8.4 — Trigonometric Identities
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Explanation
- The key step is isolating $\cos\theta$ and $\sin\theta$ from the two given expressions by simple rearrangement.
- Once you substitute, the LHS becomes $\cos^2\theta + \sin^2\theta$, and the fundamental identity (proved in §8.4 from Pythagoras' Theorem) gives the result directly.
- Examiners award marks for: (1) correct rearrangement, (2) substitution into LHS, (3) applying the identity and concluding. Show all three steps clearly.