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:

This derivation represents the equation of an ellipse centered at \((h, k)\). Whenever you see \(\cos\) and \(\sin\) in parametric form, squaring and adding is the most common technique to eliminate \(\theta\).

Answer 1

Step 1: Understanding the Concept:
We need to isolate the trigonometric terms and use the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\).
Step 2: Key Formula or Approach:
Transform the equations to find \(\cos \theta\) and \(\sin \theta\), then square and add them.
Step 3: Detailed Explanation:
Given:
\(x = h + a\cos\theta \implies x - h = a\cos\theta \implies \frac{x-h}{a} = \cos\theta\)
\(y = k + b\sin\theta \implies y - k = b\sin\theta \implies \frac{y-k}{b} = \sin\theta\)
Now, square both equations:
\[ \left(\frac{x-h}{a}\right)^2 = \cos^2\theta \]
\[ \left(\frac{y-k}{b}\right)^2 = \sin^2\theta \]
Add the two results:
\[ \left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta \]
Using 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:
Hence proved.