91Ó°ÊÓ

For each parametric equation set, we will solve one equation for \(t\) and then substitute it into the other equation, resulting in an equation in Cartesian coordinates. a. \(x = 2t^2, y = 4 + t\) From the first equation, $$t^2 = \frac{x}{2} \Rightarrow t = \pm \sqrt{\frac{x}{2}}$$ Substitute this into the second equation: $$y = 4 \pm \sqrt{\frac{x}{2}}$$ b. \(x = 2t^4, y = 4 + t^2\) From the first equation, $$t^4 = \frac{x}{2} \Rightarrow t^2 = \pm \sqrt[4]{\frac{x}{2}}$$ Substitute this into the second equation: $$y = 4 \pm \sqrt[4]{\frac{x}{2}}$$ c. \(x = 2t^{\frac{2}{3}}, y = 4 + t^{\frac{1}{3}}\) From the first equation, $$t^{\frac{2}{3}} = \frac{x}{2} \Rightarrow t = \left(\frac{x}{2}\right)^{\frac{3}{2}}$$ Substitute this into the second equation: $$y = 4 + \left(\frac{x}{2}\right)^{\frac{1}{2}}$$

Now that we have the equations in Cartesian form, let's compare them: a. \(y = 4 \pm \sqrt{\frac{x}{2}}\) b. \(y = 4 \pm \sqrt[4]{\frac{x}{2}}\) c. \(y = 4 + \left(\frac{x}{2}\right)^{\frac{1}{2}}\) We can see that none of these equations are the same, as the exponents and the signs are different in each case. Thus, the given parametric equations do not describe the same curve.