Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.3 Exercises - Page 711: 15

$$\eqalign{ & {\text{At }}\left( { - \frac{2}{{\sqrt 3 }},\frac{3}{2}} \right):{\text{ }}3\sqrt 3 x - 8y + 18 = 0 \cr & {\text{At }}\left( {0,2} \right):{\text{ }}y = 2 \cr & {\text{At }}\left( {2\sqrt 3 ,\frac{1}{2}} \right):{\text{ }}\sqrt 3 x + 8y - 10 = 0 \cr} $$

$$\eqalign{ & x = 2\cot \theta ,{\text{ }}y = 2{\sin ^2}\theta \cr & {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {2{{\sin }^2}\theta } \right]}}{{\frac{d}{{d\theta }}\left[ {2\cot \theta } \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{4\sin \theta \cos \theta }}{{ - 2{{\csc }^2}\theta }} \cr & \frac{{dy}}{{dx}} = - 2{\sin ^3}\theta \cos \theta \cr & {\text{*For }}\left( { - \frac{2}{{\sqrt 3 }},\frac{3}{2}} \right) \to \theta = \frac{{2\pi }}{3} \cr & {\text{Calculate the tangent line at the point }}\left( { - \frac{2}{{\sqrt 3 }},\frac{3}{2}} \right) \cr & {\text{Evaluate at }}\theta = \frac{{2\pi }}{3} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{{2\pi }}{3}}} = - 2{\sin ^3}\left( {\frac{{2\pi }}{3}} \right)\cos \left( {\frac{{2\pi }}{3}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{{2\pi }}{3}}} = \frac{{3\sqrt 3 }}{8} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{3}{2} = \frac{{3\sqrt 3 }}{8}\left( {x + \frac{2}{{\sqrt 3 }}} \right) \cr & y - \frac{3}{2} = \frac{{3\sqrt 3 }}{8}x + \frac{3}{4} \cr & 8y - 12 = 3\sqrt 3 x + 6 \cr & 3\sqrt 3 x - 8y + 18 = 0 \cr & \cr & {\text{*For }}\left( {0,2} \right) \to \theta = \frac{\pi }{2} \cr & {\text{Evaluate at }}\theta = \frac{\pi }{2} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{2}}} = - 2{\sin ^3}\left( {\frac{\pi }{2}} \right)\cos \left( {\frac{\pi }{2}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{2}}} = 0 \cr & {\text{Calculate the tangent line at the point }}\left( {0,2} \right) \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = 0\left( {x + \frac{2}{{\sqrt 3 }}} \right) \cr & y = 2 \cr & \cr & {\text{*For }}\left( {2\sqrt 3 ,\frac{1}{2}} \right) \to \theta = \frac{\pi }{6} \cr & {\text{Calculate the tangent line at the point }}\left( {2\sqrt 3 ,\frac{1}{2}} \right) \cr & {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{6}}} = - 2{\sin ^3}\left( {\frac{\pi }{6}} \right)\cos \left( {\frac{\pi }{6}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{6}}} = - \frac{{\sqrt 3 }}{8} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{1}{2} = - \frac{{\sqrt 3 }}{8}\left( {x + 2\sqrt 3 } \right) \cr & y - \frac{1}{2} = - \frac{{\sqrt 3 }}{8}x - \frac{3}{4} \cr & 8y - 2 = - \sqrt 3 x - 6 \cr & \sqrt 3 x + 8y - 10 = 0 \cr} $$