# Ellipse inscribed in a circle

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?

• 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, $PP_2 = a \sin E$. Jun 21, 2021 at 14:41
• Can you see why the coordinate of $P$ is $(a \cos E, 0)$? Jun 21, 2021 at 14:48
• So the line parallel to y-axis through $P$ will be $x = a \cos E$. Jun 21, 2021 at 14:49
• If you want a synthetic proof, please explain what definition of ellipse you want to consider. Jun 21, 2021 at 15:28
• @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. Jun 21, 2021 at 17:13

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.
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).
$$P_{1}P$$ = bsint(t), $$P_{2}P$$ = asin(t) Therefore, $$\frac{P_{1}P}{P_{2}P}$$ = $$\frac{b}{a}$$. Here t is just a parametre.
• 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. Jun 21, 2021 at 14:32