91Ó°ÊÓ

For trigonometric functions, using the angle \(t\) as our parameter is a common choice. To express both \(x\) and \(y\) in terms of \(t\), we need to find a relationship between \(y\) and \(t\). Let \(\cos y = a\) and \( \sin y = b\). Therefore, \(\cos^{2}y = a^{2}\) and \(\sin^{2}y = b^{2}\). Now, recall the relationship between \(\sin\) and \(\cos\): \(\sin^{2}\theta + \cos^{2}\theta = 1\). That means, \(a^{2} + b^{2} = 1\). From the given equation, \(x = 1 + \cos^{2} y - \sin^{2} y\), we can now replace \(\cos^{2}y\) and \(\sin^{2}y\) by \(a^{2}\) and \(b^{2}\) respectively. So, \(x = 1 + a^{2} - b^{2}\). Now, we have the following relationships: 1. \(a = \cos y\) 2. \(b = \sin y\) 3. \(a^{2} + b^{2} = 1\) 4. \(x = 1 + a^{2} - b^{2}\) Now we can express \(a\) and \(b\) in terms of a single parameter, \(t\). If we let \(a = \cos t\) and \(b = \sin t\), then the third equation becomes \(\cos^{2}t + \sin^{2}t = 1\). This is also true, which means our choice is valid. Now, we can express the original equation using \(a\) and \(b\) in terms of \( t\): 1. \(a = \cos t\) 2. \(b = \sin t\) 3. \(x = 1 + \cos^{2}t - \sin^{2}t\) 4. \(y = \arccos a = \arccos(\cos t) = t\) (since \(\cos y = a\))

Now, we have the following parametric equations for \(x\) and \(y\) in terms of the parameter \(t\): 1. \(x(t) = 1 + \cos^{2}t - \sin^{2}t\) 2. \(y(t) = t\)

Now that we have our parametric equations, we can use a graphing utility such as Desmos, GeoGebra, or a graphing calculator to plot the curve. For the utility, set the parameter range as continuous for \(t\). The graph of the curve \(x = 1 + \cos^{2}y - \sin^{2}y\) will be displayed, and now you have successfully graphed the curve using parametric equations and a graphing utility.