 In this video, we will provide the solution to question number 10 from the practice final exam for math 1060 for which we're given a parametric curve where x equals 3 cosine of t and y equals 4 sine of t. This would actually give us the parameterization of an ellipse, but we're asked to eliminate the parameter. So there's two ways we could approach this. The fact that we know the parametric curve is an ellipse means that we could be looking for the equation that matches the standard general equation of an ellipse. But let's suppose we didn't know that. How could we remove the parameterization just knowing the fact that we have this trigonometric parameterization? Well, the idea is to connect these together using some Pythagorean relationship. The idea here is that cosine squared of t plus sine squared of t is equal to 1. And so if you solve for cosine and sine in this situation, you get cosine t is equal to x over 3. You get sine t is equal to y over 4. Substituting that into this equation, we end up with x over 3 squared plus y over 4 squared is equal to 1. Squaring the denominators, we get x squared over 9 plus y squared over 16. This is equal to 1. And so this does give us, of course, the general, or this gives us the equation of the ellipse, which is centered at the origin, which has a horizontal radius of 3, a vertical radius of 4. But again, without even understanding these equations, the conic sections, we're able to use the Pythagorean equation to eliminate the parameter in the situation, from which case we then see that the correct answer would be choice b.