Question 967343
sin(t) = (x-3)/2


cos(t) = y/3


sin^2(t) + cos^2(t) = 1 becomes:


((x-3)/2)^2 + (y/3)^2 = 1 which becomes:


(x-3)^2 / 2^2 + y^2 / 3^2 = 1 which becomes:


(x-3)^2 / 4 + y^2 / 9 = 1


this looks like it's the equation of an ellipse.


the graph of that equation is shown below:


in the graph, 3 (x,y) pairs were shown to be on the graph of the equation.


these points were derived from the parametric equations for x and y.


this confirmed that the graph is an accurate representation of the parametric equations.


the 3 points were based on t = 0, t = 45 degrees, t = 270 degrees.


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