Question

Consider an arbitrary ellipse x 2 a 2 + y 2 b 2 = 1. (Assume a, b > 0.) (a) Find a parametrization of the ellipse using the domain t ∈ [0, 2π] such that the initial point and terminal point are both (0, b) and the ellipse is traversed exactly once in the counterclockwise direction (except that one point is hit twice). On a sketch of the ellipse, label the "compass points" of the ellipse with their t-values.

Answer 1

Answer:

α(t) = (-a*Sin(t), b*Cos (t))    where t ∈ [0, 2π]

Step-by-step explanation:

Given an arbitrary ellipse (x²/a²) + (y²/b²) = 1       (a, b > 0)

The parametrization can be as follows

x = -a*Sin(t)

y = b*Cos (t)

then

α(t) = (-a*Sin(t), b*Cos (t))    where t ∈ [0, 2π]

If  t = 0

α(0) = (-a*Sin(0), b*Cos (0)) = (0, b)

If  t = π/2

α(π/2) = (-a*Sin(π/2), b*Cos (π/2)) = (-a, 0)

If  t = π

α(π) = (-a*Sin(π), b*Cos (π)) = (0, -b)

If  t = 3π/2

α(3π/2) = (-a*Sin(3π/2), b*Cos (3π/2)) = (a, 0)

If  t = 2π

α(2π) = (-a*Sin(2π), b*Cos (2π)) = (0, b)

We can see the sketch in the pic.

Answer 2

Answer:

Check below

Step-by-step explanation:

1) Firstly let's rewrite the equation for the sake of clarity, bearing in mind (a, b >0):

$\frac{x^{2}}{a^2}+\frac{y^2}{b^2}=1$

a) To find a parametrization, to begin with we need to keep in mind this relations:

I)The general formula of the ellipse:

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

II) Parametrization:

$\left\{\begin{matrix}x(t)=acos(t)+p\\ y(t)=bsin(t)+p\end{matrix}\right.t\in[0,2\pi]$

So, for that ellipse arbitrarily chosen we have:

p=0, q=0. So plugging in we have:

$\left\{\begin{matrix}x(t)=acos(t)\\ y(t)=bsin(t)\end{matrix}\right.\:\:t\in[0,2\pi]$

For this exercise, suppose a ≠ b, and both >0

Since the Foci  have the size of the minor axis over the longer one then

$F_{1}=(-b,0), and \:F_{2}=(b,0)$

T values for F1, and F2

$y=-bsin(t)\\\frac{y}{sin(t)}=-b\frac{sin(t)}{sin(t)}\Rightarrow \\\frac{y}{sin(t)}=-b\\\\\frac{1}{sin(t)}y=-b\\ \frac{1}{sin(t)}=\frac{-b}{y} \\F2 \\y=bsin(t)\\\frac{y}{sin(t)}=b\frac{sin(t)}{sin(t)}\Rightarrow \\\frac{y}{sin(t)}=-b\\\\\frac{1}{sin(t)}y=b\\ \frac{1}{sin(t)}=\frac{b}{y}$