Ellipse's Circular Tangents

Let's solve it using coordinate geometry

Let the equation of ellipse be $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a \gt b$ . Let eccentricity of ellipse be $e$ .

Now, let the parametric coordinates of points on ellipse be $(a \cos \theta, b \sin\theta)$ .

Now, coordinates of endpoints of latus rectum are $\displaystyle \bigg(ae, \frac{b^2}{a}\bigg)$ and $\displaystyle \bigg(ae, -\frac{b^2}{a}\bigg)$ . Comparing with polar coordinates we get $\cos \theta = e \text{ and } \sin\theta = \pm\sqrt{1 - e^2}$

Now, equation of normal at point $(a \cos \theta, b \sin\theta)$ is $\displaystyle \frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$

So, equations of normals at endpoints of latus rectum are

$\frac{ax}{e} - \frac{by}{\sqrt{1-e^2}} = a^2 - b^2 \text{ and } \frac{ax}{e} + \frac{by}{\sqrt{1-e^2}} = a^2 - b^2$

Now the normals will intersect at circle's center $C(h, k)$ . Solving, we get

$h = e\frac{a^2 - b^2}{a} \text{ and } k = 0$

Now, $b^2 = a^2(1-e^2)$ , so

$\boxed{h = ae^3 \text{ and } k = 0}$

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Now, as circle touches minor axis, i.e. y-axis, so

$\text{radius of circle } = \text{x-coordinate of center}$

$\boxed{\text{radius of circle } = ae^3}$

Now, $\text{radius of circle } = \text{distance between center of circle and ends of latus rectum}$

$ae^3 = \sqrt{(ae^3 - ae)^2 + \bigg(\frac{b^2}{a}\bigg)^2}$

$ae^3 = \sqrt{(ae^3 - ae)^2 + (a(1-e^2))^2}$

$a^2e^6 = a^2e^6 - 2a^2e^4 + a^2e^2 + a^2 - 2a^2e^2 + a^2e^4$

$e^4 + e^2 - 1 = 0$

$\boxed{e^2 = \frac{\sqrt5 - 1}{2}}$

Now, area of circles $A_1 = 2×(\pi (ae^3)^2)= 2\pi a^2e^6$ and area of ellipse $A_2 = \pi ab = \pi a^2\sqrt{1-e^2}$

$\begin{aligned} \frac{A_1}{A_2} & = \frac{2\pi a^2e^6}{\pi a^2 \sqrt{1-e^2}} \\ & = \frac{2e^6}{\sqrt{1-e^2}} \\ & = \frac{2(\frac{\sqrt5 - 1}{2})^3}{\sqrt{1-\frac{\sqrt5 - 1}{2}}} \\ & = 2\frac{\sqrt5 - 2}{\sqrt{\frac{6-2\sqrt5}{4}}} \\ & = 4\frac{\sqrt5 - 2}{\sqrt5 - 1} \\ & = \boxed{\frac{4}{3 + \sqrt5}} \end{aligned}$

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So, $a = 4, b = 5, c = 3, a^2 + b^2 + c^2 = 50$

Visualisation: Area of circles is about $76.393\%$ of ellipse's area

Interestingly, the ellipse that satisfies the conditions is $\dfrac {x^2}{\varphi^2} + y^2 = 1$ , where the major axis $a = \varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio and the minor axis $b = 1$ or an eccentricity $e = \dfrac 1{\sqrt \varphi}$ (see proof below). Then the radius of the two circle is $r = e = \dfrac 1{\sqrt \varphi}$ . Then $\dfrac {A_1}{A_2} = \dfrac {2\pi r^2}{\pi a b} = \dfrac 2{\varphi^2} = \dfrac 2{\left(\frac {1+\sqrt 5}2\right)^2} = \dfrac 8{2\sqrt 5+6} = \dfrac 4{\sqrt 5 + 3}$ . Therefore $\bold a^2 + \bold b^2 + \bold c^2 = 4^2 + 5^2 + 3^2 = \boxed{50}$ . (Note that $\bold a$ and $\bold b$ are not major and minor axes.)

Proof: Let the ellipse be $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ with foci $F_1(-c,0)$ and $F_2(c,0)$ , where $c = \sqrt{a^2-b^2}$ .

Note that latus rectum is the straight line passes through a focus and perpendicular ot the major axis. Let the top end of the latus rectum through $F_2$ be $P(c, v)$ . By definition of ellipse $F_1P + F_2P = 2a$ or

$\begin{aligned} \sqrt{4c^2+v^2} + v & = 2a \\ \sqrt{4c^2+v^2} & = 2a - v \\ 4c^2 + v^2 & = 4a^2 - 4av + v^2 \\ \implies v & = \frac {a^2-c^2}a = \frac {b^2}a \end{aligned}$

The gradient at a point on the ellipse is given by $\dfrac {dy}{dx} = - \dfrac {b^2x}{a^2y}$ . At $P(c,v)$ , we have $\dfrac {dy}{dx}\bigg|_{x=c} = - \dfrac {b^2c}{a^2v} = - \dfrac ca$ . Let the radius of the circles be $r$ . Then the circle for $x \ge 0$ is $(x-r)^2 + y^2 = r^2$ and $\dfrac {dy}{dx} = - \dfrac {x-r}y$ . At $P(c,v)$ , $\dfrac {dy}{dx}\bigg|_{x=c} = - \dfrac {c-r}v$ . Since the circle and ellipse are tangent at $P$ , then

$\begin{aligned} \frac {c-r}v & = \frac ca \\ \frac {a(c-r)}{b^2} & = \frac ca \\ c - r & = \frac {b^2}{a^2}c \\ r & = c \left(1- \frac {b^2}{a^2} \right) = \frac {c^3}{a^2} \end{aligned}$

For the circle at $P$ :

$\begin{aligned} (c-r)^2 + v^2 & = r^2 \\ c^2 - 2cr + r^2 + \frac {b^4}{a^2} & = r^2 \\ c^2 - \frac {2c^4}{a^2} + \frac {b^4}{a^2} & = 0 \\ a^2c^2 - 2c^4 + b^4 & = 0 \\ a^2(a^2-b^2) - 2(a^4 - 2a^2b^2 + b^4) + b^4 & = 0 \\ a^4 - 3a^2b^2 + b^4 & = 0 \end{aligned}$

$\begin{aligned} a^2 & = \frac {3b^2 + \sqrt{9b^4-4b^4}}2 = \frac {3+\sqrt 5}2b^2 & \small \blue{\text{Note that }\varphi^2 = \left(\frac {1+\sqrt 5}2\right)^2 = \frac {3+\sqrt 5}2} \\ \implies a & = \varphi b \end{aligned}$

Putting $b=1$ , $\implies a = \varphi$ , $c = \sqrt{\varphi^2 -1} = \sqrt{\varphi +1-1} = \sqrt \varphi$ , and $r = \dfrac {c^3}{a^2} = \dfrac 1{\varphi}$ .