Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.6 Parametric Equations, Graphs, and Applications - 8.6 Exercises - Page 399: 39b

$y = \sqrt{3}~x-\frac{x^2}{36}$

From Part (a): $x = 24~t$ $y = 24\sqrt{3}~t-16~t^2$ We can find an expression for $t$ in terms of $x$: $x = 24~t$ $t = \frac{x}{24}$ We can substitute this value in the equation for $y$: $y = 24\sqrt{3}~t-16~t^2$ $y = 24\sqrt{3}~(\frac{x}{24})-16~(\frac{x}{24})^2$ $y = \sqrt{3}~x-\frac{x^2}{36}$