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The rectangular equation of the curve given parametrically by x=1+e^-t and y=1+e^t is (a) y=2-x (b) y=x (c) y=1/x (d) y=x/(x-1)
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d
\[ y-1=e^t\\ x-1 =e^{-t}\\ y-1=\frac{1}{x-1}\\ y=\frac{x}{x-1} \]
care to explain the jump from y−1=1/x−1 to the answer?
\[y-1 = \frac 1{x-1}\\ y =1+ \frac 1{x-1}\\ y= \frac {(x-1)+1} {x-1} \\ y=\frac {x} {x-1} \\\]
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