Chapter 10 - Parametric And Polar Curves; Conic Sections - 10.1 Parametric Equations; Tangent Lines And Arc Length For Parametric Curves - Exercises Set 10.1 - Page 701: 11

$$y=\dfrac{6-3x}{2}$$ $$ \text{ Straight Line }$$

We have: $x=2 \sin^2t \implies \sin^2 t=\dfrac{x}{2}$ and $y= \cos^2t \implies \cos^2 t=\dfrac{y}{3}$ Now, use the identity $$\sin^2 t+\cos^2 t=1 \dfrac{x}{2}+\dfrac{y}{3}\\ 1=\dfrac{3x+2y}{6} \\y=\dfrac{6-3x}{2}$$ This shows a straight line.