Ellipse's Circumscribed Perimeter

Lets solve it using coordinate geometry

Let equation of ellipse be $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Let tangent be drawn at point $P(a\cos{\theta}, b\sin{\theta})$

Equation of tangent at $P$ is

$\frac{x}{a} \cos{\theta} + \frac{y}{b} \sin{\theta} = 1$

It intersects the major and minor at $C(a\sec{\theta}, 0)$ and $B(0, b\cosec{\theta})$

Length of $BC$ is

$L = \sqrt{a^2\sec^2{\theta} + b^2\cosec^2{\theta}}$

To minimise $L$ , $\displaystyle \frac{dL}{d\theta} = 0$

$\frac{dL}{d\theta} = \frac{2a^2\sec^2{\theta}\tan{\theta} - 2b^2\cosec^2{\theta}\cot{\theta}}{2\sqrt{a^2\sec^2{\theta} + b^2\csc^2{\theta}}} = 0$

$\frac{b^2}{a^2} = \tan^4{\theta} \text{ or } \tan^2{\theta} = \frac{b}{a}$

$\sec{\theta} = \sqrt{\frac{a+b}{a}} \text{ and } \cosec{\theta} = \sqrt{\frac{a+b}{b}}$

Now, perimeter will be minimum when all sides are of this length $L_{min} =\sqrt{a^2\sec^2{\theta} + b^2\cosec^2{\theta}} = a + b$

So, area of quadrilateral is

$A_2 = 4×\frac12(a\sec{\theta})(b\cosec{\theta})$

$A_2 = 2ab\frac{a+b}{\sqrt{ab}}$

$\boxed{A_2 = 2\sqrt{ab}(a+b)}$

Now, Area of ellipse is $A_1 = \pi ab$

So,

$\frac{A_1}{A_2} = \frac{\pi ab}{2\sqrt{ab}(a+b)}$

$\frac{A_1}{A_2} = \frac{\pi}{2(\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}})}$

Now, $\displaystyle e^2 = 1-\frac{b^2}{a^2}$ or $\displaystyle \sqrt{\frac{b}{a}} = \sqrt[4]{1-e^2}$

Therefore,

$\boxed{\frac{A_1}{A_2} = \frac{\pi}{2(\sqrt[4]{1-e^2}+\frac{1}{\sqrt[4]{1-e^2}})}}$

Putting $e = 0.85$ , we get

$\frac{A_1}{A_2} = \frac{\pi}{2(\sqrt[4]{1-0.85^2}+\frac{1}{\sqrt[4]{1-0.85^2}})}$

$\frac{A_1}{A_2} = 0.74672$

$\color{#3D99F6}{\boxed{\frac{A_1}{A_2}×100 = 74.672}}$

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Therefore, area of ellipse $A_1$ is $\boxed{74.672\%}$ of area of quadrilateral $A_2$ .

Fun fact(or bonus question): minimum area of quadrilateral gives area of ellipse as $25\pi = 78.538\%$ of quadrilateral's area. Can you prove it?

By differentiating the ellipse equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get $\frac{2xdx}{a^2}+\frac{2ydy}{b^2}=0$ and hence $\frac{dy}{dx}=-\frac{b^2}{a^2}\frac{x}{y}$ Now suppose the line segment in the first quadrant is tangent at the point (x,y), then it will intersect the y-axis at $x+c$ and the x-axis at $y+d$ where $c=-y\frac{dx}{dy}=\frac{a^2y^2}{b^2x}$ and $d=-x\frac{dy}{dx}=\frac{b^2x^2}{a^2y}$ .

If s is the length of our line segment then we have $s^2=(y+\frac{b^2x^2}{a^2y})^2 +(x+\frac{a^2y^2}{b^2x})^2$ which is rewritten as $s^2=(\frac{1}{y^2a^4}+\frac{1}{x^2b^4})(a^2y^2+b^2x^2)^2$

Using the ellipse equation we can substitute $a^2y^2=b^2(a^2-x^2)$ , so that $s^2=(\frac{1}{a^2b^2(a^2-x^2)}+\frac{1}{x^2b^4})(b^2a^2)^2$

At minimum value for s we must have that $\frac{ds^2}{dx}=0$ , from which it follows that $a^2(a^2-x^2)=b^2x^2$

Now we use the relationship $b^2=(1-e^2)a^2$ to get $e^2x^4-2a^2x^2+a^4=0$ which is a quadratic in $x^2$ that has roots $x^2=\frac{a^2}{e^2}(1\pm\sqrt{1-e^2})$ . We discard the root with $|x|>a$ , so that $\frac{x}{a}=\frac{1}{e}\sqrt{1-\sqrt{1-e^2}}$ Hence we also find $\frac{y}{b}=\frac{1}{e}\sqrt{e^2-1+\sqrt{1-e^2}}$ $\frac{c}{a}=\frac{a}{x}\frac{y^2}{b^2}=\frac{e^2-1+\sqrt{1-e^2}}{e\sqrt{1-\sqrt{1-e^2}}}$ $\frac{d}{b}=\frac{b}{y}\frac{x^2}{a^2}=\frac{1-\sqrt{1-e^2}}{e\sqrt{e^2-1+\sqrt{1-e^2}}}$

The area ratio can now be written as $\frac{πab}{4× \frac{1}{2}(x+c)(y+d)}=\frac{π}{2}\frac{a}{x+c}\frac{b}{y+d} =\frac{π}{2}\frac{\sqrt{1-\sqrt{1-e^2}}}{e}\frac{\sqrt{e^2-1+\sqrt{1-e^2}}}{e} =\frac{π}{2}\frac{\sqrt[4]{1-e^2}}{1+\sqrt{1-e^2}}$

Let solve this problem for a general case of $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ , where $a$ and $b$ are the major semiaxis and minor semiaxis respectively. The gradient of a point on an ellipse is given by differentiating the equation on both sides with respect to $x$ , $\dfrac {2x}{a^2} + \dfrac {2y}{b^2} \cdot \dfrac {dy}{dx} = 0 \implies \dfrac {dy}{dx} = - \dfrac {b^2x}{a^2y}$ .

Let a side of the quadrilateral $AB$ tangent to the ellipse at $P(u,v)$ . Then we have $\dfrac {y-v}{x-u} = - \dfrac {b^2u}{a^2v}$ . Let $A(0, y_0)$ and $B(x_0, 0)$ . Then $\dfrac {y_0-v}{-u} = - \dfrac {b^2u}{a^2v} \implies y_0 = \dfrac {b^2u^2}{a^2v} + v = \dfrac {b^2}v \left(\cancel{\dfrac {u^2}{a^2} + \dfrac {v^2}{b^2}}^\red{\ 1} \right) = \dfrac {b^2}v$ . Similarly, $x_0 = \dfrac {a^2}u$ . Let $u = a\cos \theta$ and $v=b\sin \theta$ . Then $x_0 = \dfrac a{\cos \theta}$ and $y_0 = \dfrac b{\sin \theta}$ .

Let the length $AB= \ell$ . Then the perimeter of quadrilateral is $4\ell$ . The perimeter is shortest when $\ell$ is shortest or $\ell^2$ is the minimum. Then we have:

$\begin{aligned} \ell^2 & = x_0^2 + y_0^2 = \frac {a^2}{\cos^2 \theta} + \frac {b^2}{\sin^2 \theta} & \small \blue{\text{Differentiate both sides w.r.t. }\theta} \\ \frac {d \ell^2}{d \theta} & = \frac {a^2\sin \theta}{\cos^3\theta} - \frac {b^2\cos \theta}{\sin^3 \theta} & \small \blue{\text{Putting }\frac {d\ell^2}{d\theta} = 0} \\ \frac {a^2\sin \theta}{\cos^3\theta} & = \frac {b^2\cos \theta}{\sin^3 \theta} \\ \tan^4 \theta & = \frac {b^2}{a^2} \\ \implies \tan \theta & = \sqrt{\frac ba} & \small \blue{\implies \sin \theta = \sqrt{\frac b{a+b}}, \cos \theta = \sqrt{\frac a{a+b}}} \end{aligned}$

Since $\dfrac {d^2 \ell^2}{dx^2} > 0$ , when $\theta = \tan^{-1} \sqrt{\dfrac ba}$ , therefore $\ell$ is minimum when $\theta = \tan^{-1} \sqrt{\dfrac ba}$ . Then $x_0 = \dfrac a{\cos \theta} = \sqrt{a(a+b)}$ and $y_0 = \dfrac b{\sin \theta} = \sqrt{b(a+b)}$ . And $A_2 = 2x_0y_0 = 2(a+b)\sqrt{ab}$ . Since $A_1 = \pi a b$ , then $\dfrac {A_1}{A_2} \times 100 = \dfrac {50\pi\sqrt{ab}}{a+b}$ .

For eccentricity $e = 0.85 = \dfrac {17}{20} = \sqrt{1-\dfrac {b^2}{a^2}}$ .. $\implies \dfrac {b^2}{a^2} = 1 - e^2 = \dfrac {111}{400}$ , $\implies a = 20$ , $b = \sqrt{111}$ and $\dfrac {A_1}{A_2} \times 100 = \dfrac {50\pi\sqrt{20\sqrt{111}}}{20+\sqrt{111}} \approx \boxed{74.7}$ .