Question

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

Hint:

When proving trig identities, keep an eye on the RHS to decide whether to convert everything to \(\sin/\cos\) early or use identities like \(1 - \sec^2 A = -\tan^2 A\).

Answer 1

Step 1: Understanding the Concept:
We need to isolate the trigonometric terms (\(\sin \theta\) and \(\cos \theta\)) and use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\).
Step 2: Key Formula or Approach:
Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
Step 3: Detailed Explanation:
1. From \(x = h + a \cos \theta\), we get: \(x - h = a \cos \theta \implies \frac{x - h}{a} = \cos \theta\). 2. From \(y = k + b \sin \theta\), we get: \(y - k = b \sin \theta \implies \frac{y - k}{b} = \sin \theta\). 3. Squaring and adding both equations: \[ \left( \frac{x - h}{a} \right)^2 + \left( \frac{y - k}{b} \right)^2 = \cos^2 \theta + \sin^2 \theta \] 4. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\): \[ \left( \frac{x - h}{a} \right)^2 + \left( \frac{y - k}{b} \right)^2 = 1 \]
Step 4: Final Answer:
LHS = RHS. Hence proved.