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\noindent \textbf{Problem}: ($\#$28, pg. 25) Convert the parametric equations into equations of a curve in rectangular form:
\[\left\{ \begin{array}{ll}
x = \sqrt{t}, & \quad (i) \\
y=2t+4, & \quad (ii)
\end{array} \right.\]
\textit{Solution}: Solve $(ii)$ for $t$ to get $t = \dfrac{y-4}{2}$. Plug this $t$ into $(i)$ to get $x = \sqrt{ \dfrac{y-4}{2} }$. This completes the problem.
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\noindent \textbf{NOTE} You can also do this a different way: solve $(i)$ for $t$ to get $t = x^2$ and plug this into $(ii)$ to get $y=2x^2+4$. This answer is equivalent! To see that, solve it for $x$. To do this we get to $x^2 = \dfrac{y-4}{2}$ and when we take square roots we must consider $\pm$: $x = \pm \sqrt{ \dfrac{y-4}{2} }$ and we must take the $+$ solution here, giving us the equation we got the first time.
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