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If $x = h + a \cos \theta$, $y = k + b \sin \theta$, then prove that : $(\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = 1$
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$x = h + a \cos \theta \Rightarrow \frac{x-h}{a} = \cos \theta$ (I) (1 Mark)
$y = k + b \sin \theta \Rightarrow \frac{y-k}{b} = \sin \theta$ (II) (1 Mark)
$\therefore LHS = (\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = \cos^2\theta + \sin^2\theta = 1 = RHS$ (III) (1 Mark)
$y = k + b \sin \theta \Rightarrow \frac{y-k}{b} = \sin \theta$ (II) (1 Mark)
$\therefore LHS = (\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = \cos^2\theta + \sin^2\theta = 1 = RHS$ (III) (1 Mark)