# Video: Finding the Cartesian Equation of a Curve That Is Defined by Two Parametric Equations

Find the Cartesian equation of the curve defined by the parametric equations π₯ = 2 + cos π‘ and π¦ = 4 cos 2π‘.

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### Video Transcript

Find the Cartesian equation of the curve defined by the parametric equations π₯ equals two plus cos π‘ and π¦ equals four cos of two π‘.

Remember, a Cartesian equation is one which contains only the variables π₯ and π¦. So weβre going to need to find a way to eliminate our third variable π‘ from our parametric equations. And at first glance, it doesnβt seem to be a nice way to do so. But we can begin by recalling some trigonometric identities. We have cos of two π‘ in our second parametric equation. And we know that cos of two π‘ is equal to two times cos squared π‘ minus one. This means we can rewrite our equation for π¦ as four times two cos squared π‘ minus one.

Next, weβll look to rearrange our equation for π₯ to make cos of π‘ the subject. Once weβve done that, weβll be able to find an expression for cos squared π‘ in terms of π₯. We can subtract two from both sides. And we see that π₯ minus two equals cos of π‘. Then, by squaring both sides of this equation, we find that cos squared π‘ is equal to π₯ minus two all squared. And so, weβre now able to replace cos squared π‘ with π₯ minus two squared. That gives us π¦ equals four times two times π₯ minus two all squared minus one.

We distribute this first pair of parentheses. And we find that π₯ minus two all squared is equal to π₯ squared minus four π₯ plus four. We distribute again by multiplying each of these terms by two and then simplifying: eight minus one is seven. Well, finally, we distribute one more time by multiplying each term of two π₯ squared minus eight π₯ plus seven by four. And we find the Cartesian equation of the curve defined by parametric equations π₯ equals two plus cos π‘ and π¦ equals four cos of two π‘ is π¦ equals eight π₯ squared minus 32π₯ plus 28.