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Please, help me with the following problem:

Starting from the figure below, ellipse inscribed in a circle

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?

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    $\begingroup$ Equation of line $PP_1$ is $x = a \cos E \ $. Plug it in the equation of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and find $y$. That is $PP_1$ and it comes to $b \sin E$. Also, $PP_2 = a \sin E$. $\endgroup$
    – Math Lover
    Jun 21, 2021 at 14:41
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    $\begingroup$ Can you see why the coordinate of $P$ is $(a \cos E, 0)$? $\endgroup$
    – Math Lover
    Jun 21, 2021 at 14:48
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    $\begingroup$ So the line parallel to y-axis through $P$ will be $x = a \cos E$. $\endgroup$
    – Math Lover
    Jun 21, 2021 at 14:49
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    $\begingroup$ If you want a synthetic proof, please explain what definition of ellipse you want to consider. $\endgroup$ Jun 21, 2021 at 15:28
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    $\begingroup$ @Augustin Yes, but how do you define an ellipse? By the sum of distances to foci? By the focus-directrix property? Can we take as given, for instance, that $(PO/a)^2+(P_1P/b)^2=1$? You must give some more details. $\endgroup$ Jun 21, 2021 at 17:13

2 Answers 2

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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.

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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.

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  • $\begingroup$ Thank you, @Ayus Das, but the point $P_{2}$ should not have coordinates $(a\cos t, a\sin t)$? I apologize if I'm wrong. $\endgroup$
    – Augustin
    Jun 21, 2021 at 14:32
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    $\begingroup$ @Augustin Yes, that was a typo, I presume. $\endgroup$ Jun 21, 2021 at 15:27

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