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):
[tex]\frac{x^{2}}{a^2}+\frac{y^2}{b^2}=1[/tex]
a) To find a parametrization, to begin with we need to keep in mind this relations:
I)The general formula of the ellipse:
[tex]\\\frac{(x-p)^2}{a^2}+\frac{(y-q)^2}{b^2}=1[/tex]
II) Parametrization:
[tex]\left\{\begin{matrix}x(t)=acos(t)+p\\ y(t)=bsin(t)+p\end{matrix}\right.t\in[0,2\pi][/tex]
So, for that ellipse arbitrarily chosen we have:
p=0, q=0. So plugging in we have:
[tex]\left\{\begin{matrix}x(t)=acos(t)\\ y(t)=bsin(t)\end{matrix}\right.\:\:t\in[0,2\pi]\\[/tex]
For this exercise, suppose a ≠ b, and both >0
Since the Foci have the size of the minor axis over the longer one then
[tex]F_{1}=(-b,0), and \:F_{2}=(b,0)\\[/tex]
T values for F1, and F2
[tex]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} \\[/tex]
