Question

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.)

Answer 1

Answer and Step-by-step explanation:

The computation of points on the ellipse is shown below:-

Distance between any point on the ellipse

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

To minimize

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

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

t            $d^{2} = (3\ cos t - \frac{4}{3})^{2} + sin^{2}t$

0           $\frac{25}{9}$<u />

<u />$\pi$           $\frac{169}{9}$

$2\pi$         $\frac{25}{9}$

$\frac{\pi}{3}$          $\frac{7}{9}$

$\frac{5\pi}{3}$         $\frac{7}{9}$

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

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