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I have the equation $16x^2+25y^2=400$, and the parametric equation $(x,y)=(5\cos t, 4\sin t)$.

If I plug in the parametric equation into the first equation, I end up with the trigonometric identity $\cos^2 t+ \sin^2 t= 1$. How does this identity show that my non-parametric equation, when graphed, will result in an ellipse?

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The pair of parametric equations $x = 5\cos t, y = 4\sin t$ in fact corresponds to the ellipse described by the Cartesian equation $16x^2 + 25y^2 = 400$. Substituting the first equation into the second would naturally give you an identity (like you got). However, to properly prove this, it's better to manipulate one form into another.

To prove this, manipulate the parametric equations:

$x^2 = 25\cos^2 t$

$\frac{1}{25}x^2 = \cos^2 t$

and similarly,

$\frac{1}{16}y^2 = \sin^2 t$

Add those up:

$\frac{1}{25}x^2 + \frac{1}{16}y^2 = \cos^2 t + \sin^2 t$

And use the trigonometric identity $\cos^2 t + \sin^2 t = 1$ to get:

$\frac{1}{25}x^2 + \frac{1}{16}y^2 = 1$

Rearranging,

$16x^2 + 25y^2 = 400$, which is exactly the Cartesian equation you expect.

You can see this is an ellipse by recognising the earlier derived canonical form:

$(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$, where $a = 5$ and $b = 4$.

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  • $\begingroup$ Thank you @cr3 for the edit, it's correct. :) $\endgroup$
    – Deepak
    Jan 6 '16 at 5:18
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What you did does not show you it is an ellipse. It shows that conversion from direct function relation to 2 parameter form is successful.

Your knowledge of ellipse form alone lets you recognize the form..

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