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

Given:
\[ x = h + a\cos\theta,\qquad y = k + b\sin\theta \]

To Prove:
\[ \left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1 \]

Step 1: Rearranging the expressions
From the first equation: \[ x - h = a\cos\theta \] Divide both sides by \(a\): \[ \frac{x - h}{a} = \cos\theta \]

From the second equation: \[ y - k = b\sin\theta \] Divide both sides by \(b\): \[ \frac{y - k}{b} = \sin\theta \]

Step 2: Square both equations
\[ \left(\frac{x - h}{a}\right)^2 = \cos^2\theta \] \[ \left(\frac{y - k}{b}\right)^2 = \sin^2\theta \]

Step 3: Add the two results
\[ \left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = \cos^2\theta + \sin^2\theta \]

Step 4: Use the Pythagorean identity
\[ \cos^2\theta + \sin^2\theta = 1 \]

Final Result:
\[ \left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1 \] Hence proved.