Find the point(s) on the ellipse x = 3 cost, y = sin t, 0 less than or equal to t less than or equal to 2pi closest to the point(4/3,0) (Hint: Minimize the square of the distance as a function of t.) The point(s) on the ellipse closest to the given point is(are) . (Type ordered pairs. Use a comma to separate answers as needed.)

[tex](3 cos t, sin t) and (\frac{4}{3},0) is\\\\ d = \sqrt{(3 cos\ t - \frac{4}{3}^2) } + (sin\ t - 0)^2\\\\ d^2 = (3 cos\ t - \frac{4}{3})^2 + sin^2 t[/tex]

To minimize

[tex]d^2, set\ f' (t) = 0\\\\2(3cos\ t - \frac{x=4}{3} ).3(-sin\ t) + 2sin\ t\ cos\ t = 0\\\\ 8 sin\ t - 16 sin\ t\ cos\ t = 0\\\\ 8 sin\ t (1 - 2 cos\ t) = 0\\\\ sin\ t = 0, cos\ t = \frac{1}{2} \\\\ t= 0, \ 0, \pi,2\pi,\frac{\pi}{3} , \frac{5\pi}{3}[/tex]

Now we create a table by applying the critical points which are shown below:

t [tex]d^{2} = (3\ cos t - \frac{4}{3})^{2} + sin^{2}t[/tex]

0 [tex]\frac{25}{9}[/tex]

[tex]\pi[/tex] [tex]\frac{169}{9}[/tex]

[tex]2\pi[/tex] [tex]\frac{25}{9}[/tex]

[tex]\frac{\pi}{3}[/tex] [tex]\frac{7}{9}[/tex]

[tex]\frac{5\pi}{3}[/tex] [tex]\frac{7}{9}[/tex]

When t = [tex]\frac{\pi}{3}[/tex], x is [tex]\frac{3}{2}[/tex] and y is [tex]\frac{\sqrt{3} }{2}[/tex]. So, the required points are [tex](\frac{3}{2},\frac{\sqrt{3} }{2})[/tex]

When t = [tex]\frac{5\pi}{3}[/tex], x is [tex]\frac{3}{2}[/tex] and y is [tex]\frac{-\sqrt{3} }{2}[/tex]. So, the required points are [tex](\frac{3}{2},\frac{-\sqrt{3} }{2})[/tex]