(A) If $x = h + a\cos heta$, $y = k + b\sin heta$, then prove tha...

Question

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

Accepted Answer

Given: $x = h+a\cos	heta \Rightarrow \cos	heta = \frac{x-h}{a}$. Also: $y = k+b\sin	heta \Rightarrow \sin	heta = \frac{y-k}{b}$. Squaring and adding both: $\cos^2	heta+\sin^2	heta = \left(\frac{x-h}{a}ight)^2+\left(\frac{y-k}{b}ight)^2$. Since $\sin^2	heta+\cos^2	heta=1$ (identity): $\left(\frac{x-h}{a}ight)^2+\left(\frac{y-k}{b}ight)^2 = 1$ (Proved)

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