Ellipse inscribed in a circle

Question

Please, help me with the following problem:

Starting from the figure below,

we know:

$FP=OP-OF=a\cos E-ae$

and from the right triangle $OP_{2}P$, I determined

$P_{2}P = a\sin E$.

I would like to ask you: how can I prove that $P_{1}P/P_{2}P=b/a$ (where a is semi-major axis and b is semi-minor axis of ellipse)?

Can a demonstration be made by means of synthetic geometry or without equation of ellipse?

Answer 1

Actually that circle is called the auxiliary circle of the ellipse and has many nice properties. It can be used to parameterize the ellipse $P_1=(a \cos x,b \sin x)$, x being the polar angle of $P_2$ from here it follows the result you asked. And the parameterization follows from standard equation of ellipse, i.e., $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ after plugging in the x coordinate.

Answer 2

Consider $P_{1}$(acos(t), bsin(t)), since its on the ellipse, $P_{2}$(acos(t), bsin(t)) since its on the circle. P will have co-ordinates (acos(t), 0).

Using distance formula:

$P_{1}P$ = bsint(t), $P_{2}P$ = asin(t) Therefore, $\frac{P_{1}P}{P_{2}P}$ = $\frac{b}{a}$. Here t is just a parametre.