Question 375015
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Given


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ a\cos(t)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ b\sin(t)]


Square both parametric equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ a^2\cos^2(t)\ \Rightarrow\ \ \frac{x^2}{a^2}\ =\ \cos^2(t)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ = b^2\sin^2(t)\ \Rightarrow\ \ \frac{y^2}{b^2}\ =\ \sin^2(t)]


Recall the Pythagorean Identity


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y^2}{b^2}\ =\ 1\ -\ \cos^2(t)]


Then substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y^2}{b^2}\ =\ 1\ -\ \frac{x^2}{a^2}]


And finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{a^2}\ +\ \frac{y^2}{b^2}\ =\ 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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