Question

Describe the motion of a particle with position (x, y) as t varies in the given interval. (For each answer, enter an ordered pair of the form x, y.) x = 4 + sin(t), y = 1 + 5 cos(t), π/2 ≤ t ≤ 2π The motion of the particle takes place on an ellipse centered at (x, y) = . As t goes from π/2 to 2π, the particle starts at the point (x, y) = and moves clockwise three-fourths of the way around the ellipse to (x, y) = .

Answer 1

Answer:

Required center of the ellips is (4,1) and it starts from (0,1) to tyhe three fourth position (5=2\sqrt{6},1).

Step-by-step explanation:

Given, parametric equations are,

$x=4=\sin t, y=1+5\cos t$   where $\frac{\pi}{2}\leq t \leq 2\pi$

by solving this we get,

$\sin t=x-4, \cos t=\frac{y-1}{5}$

$\sin^2 t+\cos^2 t=1$

$\implies \frac{(x-4)^2}{1}+\frac{(y-1)^2}{5^2}=1$

Which represents an ellipse.

To find center, one vertices hat is where from the particle starts, and one forth position that is the one foci farthese from initial position of starting say $c_2$.

Consider the general equation of an ellipse,

$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$

where (h,k) is center and a is the distance from center to vertex.

In present case (h,k)=(4,1) and a=5, therefore the particle starts from (0,1) point which is one vertex.

And if c be the distance from foci to center then,

$c^2=a^2-b^2\implies c=2\sqrt{6}$

Hence two foci are (5-2\sqrt{6},1) and at the three fourth position another foci is $c_2=(5+2\sqrt{6},1)$.