How to find Cartesian equation!!!!
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How to find Cartesian equation!!!!

[From: ] [author: ] [Date: 11-08-19] [Hit: ]
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find cartesian equation for a curve that contains the parametrized curve.
x= 5 cos t ; y= 2 sin t, 0 _< t _<2pi

I know your supposed to use x^2 + y^2 but where do i go from there???

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x = 5cost
x/5 = cost

y = 2sint
y/2 = sint

Then:
(x/5)² + (y/2)²
= (cost)² + (sint)²
= cos²t + sin²t
= 1 ...................... Use the pythagorean identity which says sin²x + cos²x = 1

So we have:
(x/5)² + (y/2)² = 1
(x²/5²) + (y²/2²) = 1

Note that this is in standard form for an ellipse.

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What you are looking for is an equation involving only x and y, no t. What you will get is an equation of the form (x/a)^2 + (y/b)^2 = 1. This is a general equation for an ellipse with centre at the origin, where a and b are the radii along the x-axis and y-axis respectively.

That we want this, we can guess from "being supposed to use x^2 + y^2," but a bit more formally, any curve x= c cos t, y = d sin t, will yield an ellipse. both the x- and y-coordinates will swing back and forth periodically, forming an ellipse, try drawing it and you will see. It can also be seen by knowing that x=cos t, y = sin t describes the unit circle. Multiplying the sine and cosine will only make the circle fatter/thinner and taller/shorter, creating an ellipse.

Knowing what needs to be done, how can we do it? Hopefully you know that (sin t)^2 + (cos t)^2 = 1. This is what we'll use. We rewrite the parametrization as x/5 = cos t, y/2 = sin t. Then (sin t)^2 + (cos t)^2 = (x/5)^2 + (y/2)^2 = 1, and we have our equation!
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