Highly Elliptical Area

Let's solve using coordinate geometry

Let the perpendicular sides be coordinate axes and $y = mx + c$ be hypotenuse.

So, if coordinates of center of ellipse are $a, b$ , it's lengths of are also semi-major and semi-minor axes are $a, b$ in any order.

So, equation of ellipse is

$\frac{(x-a)^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$

$b^2x^2 + a^2y^2 - 2ab^2x - 2a^2by + a^2b^2 = 0$

Putting $y = mx + c$ in equation

$b^2x^2 + a^2(mx + c)^2 - 2ab^2x - 2a^2b(mx + c) + a^2b^2 = 0$

$x^2(b^2 + a^2m^2) + x(2mca^2 - 2ab^2 - 2a^2bm) + (a^2c^2 + a^2b^2 - 2a^2bc) = 0$

As line touches the ellipse, do Discriminant $D = 0$

$(2mca^2 - 2ab^2 - 2a^2bm)^2 - 4(b^2 + a^2m^2)(a^2c^2 + a^2b^2 - 2a^2bc) = 0$

Solving and sumplifying above equation, we get

$\boxed{a = \frac{c^2 - 2bc}{2m(b-c)}}$

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Now, let's minimise area of ellipse

$A = \pi ab$

$A = \pi \frac{c^2 - 2bc}{2m(b-c)}b$

$A = \frac{\pi c}{2m} \frac{cb - 2b^2}{(b-c)}$

For minimising area,

$\frac{dA}{db} = 0$

$\frac{\pi c}{2m} \frac{(b-c)(c-4b) - (bc-2b^2)}{(b-c)^2} = 0$

$2b^2 - 4bc + c^2 = 0$

Solving and taking smaller value as solution

$\boxed{b = c\bigg(1 - \frac{1}{\sqrt2}\bigg)}$

Putting value of $b$ in area expression

$A = \frac{\pi c}{2m} b\frac{c - 2b}{(b-c)}$

$A_1 = \frac{\pi c}{2m} c\bigg(1 - \frac{1}{\sqrt2}\bigg)\frac{c - 2c(1 - \frac{1}{\sqrt2})}{(c(1 - \frac{1}{\sqrt2})-c)}$

$\boxed{A_1 = -\frac{\pi c^2}{2m}(3-\sqrt8)}$

Now, Finding area of triangle

$A_2 = \frac12\frac{-c}{m}c$ $\boxed{A_2 = -\frac{c^2}{2m}}$

Therefore

$\color{#3D99F6}{\boxed{\frac{A_1}{A_2} = \pi(3 - \sqrt8)}}$

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So, $a = 3, b = 8, a^2 + b^2 = 73$

Visualisation: Maximum area of ellipse is about $53.9\%$ of triangle's area.

We can stretch and skew the whole diagram and still preserve the ratio of areas, so we can make $\triangle ABC$ an isosceles right triangle with unit legs and an area of $A_2 = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$ . The ellipse will reach its maximum when it is a circle (see edit below), which also happens to be the incircle:

The radius of the incircle is $r = \frac{A_2}{s} = \frac{\frac{1}{2}}{\frac{1}{2}(1 + 1 + \sqrt{2}} = \frac{1}{2}(2 - \sqrt{2})$ , so the area of the ellipse (or in this case circle) is $A_1 = \pi (\frac{1}{2}(2 - \sqrt{2}))^2 = \frac{\pi}{2}(3 - 2\sqrt{2})$ .

That means the ratio is $\frac{A_1}{A_2} = \frac{\frac{\pi}{2}(3 - 2\sqrt{2})}{\frac{1}{2}} = \pi(3 - \sqrt{8})$ . Therefore, $a = 3$ , $b = 8$ , and $a^2 + b^2 = \boxed{73}$ .

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Edit:

$AC$ is on $y = -x + 1$ and the ellipse tangent to $AB$ and $BC$ has an equation of $\frac{(x - a)^2}{a^2} + \frac{(x - b)^2}{b^2} = 1$ . Let $(k, -k + 1)$ on $AC$ also be tangent to the ellipse. Then using the derivative of the equation of the ellipse, $\frac{2(x - a)}{a^2} + \frac{2(y - b)\frac{dy}{dx}}{b^2} = 0$ or $\frac{2(k - a)}{a^2} + \frac{2(-k + 1 - b)(-1)}{b^2} = 0$ or $(-k + 1 - b)a^2 = (k - a)b^2$ , and since $(k, -k + 1)$ is on the ellipse, $\frac{(k - a)^2}{a^2} + \frac{(-k + 1 - b)^2}{b^2} = 1$ . These two equations solve to $a = \frac{2k + 1 - \sqrt{-4k^2 + 4k + 1}}{4k}$ and $b = \frac{2k - 3 + \sqrt{-4k^2 + 4k + 1}}{4k - 4}$ .

The area of the ellipse is $A = \pi ab = \pi \cdot \frac{2k + 1 - \sqrt{-4k^2 + 4k + 1}}{4k} \cdot \frac{2k - 3 + \sqrt{-4k^2 + 4k + 1}}{4k - 4} = \pi \cdot \frac{2k^2 - 2k - 1 + \sqrt{-4k^2 + 4k + 1}}{4k(k - 1)}$ . Its derivative is $A' = \frac{\pi(2k - 1)(2k^2 - 2k - 1 + \sqrt{-4k^2 + 4k + 1}}{4(k - 1)^2k^2\sqrt{-4k^2 + 4k + 1}}$ , and equals zero only at $k = \frac{1}{2}$ . Since $A$ is concave down, this is the only maximum point, so there is only one ellipse with a maximum area when $k = \frac{1}{2}$ , which calculates to $A = \frac{\pi}{2}(3 - 2\sqrt{2})$ .

Let the ellipse be $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ , where $a$ and $b$ are the major semiaxis and minor semiaxis respectively, and the right triangle be $\triangle ABC$ . To find the maximum value of $\dfrac {A_1}{A_2}$ is same as finding the minimum value of $A_2$ with the a fixed $A_1$ or fixed $a$ and $b$ . Since $\triangle OQR$ is similar to $\triangle ABC$ , we can consider $\triangle OQR$ .

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 the hypotenuse $QP$ tangent to the ellipse at $P(u,v)$ . Then we have $\dfrac {y-v}{x-u} = - \dfrac {b^2u}{a^2v}$ . Let $Q(0, y_0)$ and $R(x_0, 0)$ . 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 area of $\triangle OQR$ be $A$ . Then $A = \dfrac {x_0y_0}2 = \dfrac {ab}{2\sin \theta \cos \theta} = \dfrac {ab}{\sin (2\theta)}$ . $\implies A_{\min} = ab$ , when $\sin (2\theta) = 1$ , or $\theta = \dfrac \pi 4$ , or $x_0 = \sqrt 2 a$ and $y_0 = \sqrt 2 b$ .

Let $A(-a, y_1$ and $B(x_1, -b)$ . By similar triangle $\dfrac {y_1}{x_0+a} = \dfrac {y_0}{x_0} \implies \dfrac {y_1}{\sqrt 2 a +a} = \dfrac ba \implies y_1 = (1+\sqrt 2)b$ . Similarly, $x_1 = (1+\sqrt 2)a$ . Then $CA = y_1 + b = (2+\sqrt 2)b$ and $BC = (2+\sqrt 2)a$ . Then $A_{2\min} = \dfrac {(2+\sqrt 2)^2ab}2 = (3+2\sqrt 2)ab$ . Then we have:

$\frac {A_1}{A_{2\min}} = \frac {\pi ab}{(3+2\sqrt 2)ab} = \frac {\pi(3-2\sqrt 2)}{9-8} = \pi(3-\sqrt 8)$

Then $\bold a^2 + \bold b^2 = 3^2 + 8^2 = \boxed{73}$ (here $\bold a$ and $\bold b$ are not the semiaxes).