Inscribed Ellipse

Solving this problem (at least, the way I did it) leads to some slightly messy algebra; the key steps are as follow:

The idea is to use the general equation for an ellipse; we'll plug in the info we have, work out the coefficients, and then work out the axes.

The general equation whose coefficients we'll find is $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ .

Firstly, recentre the ellipse at the origin, so that the tangent lines are $y=\pm 13$ and $x=\pm \frac{37}{2}$ . This simplifies the equation and makes $D=E=0$ , and $F=a^2 b^2$ , where $a,b$ are the semi-major and semi-minor axes.

Secondly, we have $\sin{\theta}=\frac{1}{\sqrt5}$ and $\cos{\theta}=\frac{2}{\sqrt5}$ . This (plus the info in the link above) gives the relations

$A = \frac{a^2}{5} + \frac{4b^2}{5}$

$B = \frac{4(b^2-a^2)}{5}$

$C = \frac{4a^2}{5} + \frac{b^2}{5}$

Plug in $y=13$ . This gives a quadratic in $x$ . Since this line is a tangent to the ellipse, the quadratic has just one solution; so its discriminant is zero. Working the algebra through and substituting in we get $a^2 + 4b^2 = 845$

Similarly, plugging in $x=\frac{37}{2}$ , we get $4a^2 + b^2 = \frac{6845}{4}$ .

Solving these gives $a^2=400$ , or $a=\boxed{20}$ .

Let $A = 37, B = 26$ , then from symmetry, the center of the ellipse is at $\mathbf{r}_C = (A/2, B/2)$ . The equation of the ellipse is,

$(\mathbf{r} - \mathbf{r}_C)^T \mathbf{R D R}^T (\mathbf{r} - \mathbf{r}_C) = 1$

where $\mathbf{R}$ is the rotation matrix, given by,

$R = R(\theta) = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$

and $\mathbf{D} = \text{diag} \{ 1/a^2 , 1/b^2 \}$ with $a, b$ being the semi-major and semi-minor axes lengths.

The gradient (the normal to the curve) is

$\nabla = 2 \mathbf{R D R}^T (\mathbf{r} - \mathbf{r}_C)$

At the right tangency point $\mathbf{r}_1$ , the gradient is a positive multiple of $\mathbf{i}$ (the unit vector along the positive x-axis), hence

$\mathbf{R D R}^T (\mathbf{r}_1 - \mathbf{r}_C) = \alpha \mathbf{i}$

from which,

$(\mathbf{r}_1 - \mathbf{r}_C) = \alpha \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i}$

Plugging this into the ellipse equation, gives

$\alpha = \dfrac{1}{ \sqrt{ \mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} } }$

Now, the x-component of $(\mathbf{r}_1 - \mathbf{r}_C)$ at the tangency point is just $\dfrac{A}{2}$ , so

$\dfrac{A}{2} = \alpha \mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} = \sqrt{ \mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} }$

A similar equation can be developed for the upper tangency point, namely,

$\dfrac{B}{2} = \sqrt{ \mathbf{j}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{j} }$

Now $\mathbf{D}^{-1} = \text{diag} \{ a^2, b^2 \}$ , so

$\mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} = a^2 \cos^2 \theta + b^2 \sin^2 \theta$

and

$\mathbf{j}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{j} = a^2 \sin^2 \theta + b^2 \cos^2 \theta$

And we end up with the following linear system in $a^2$ and $b^2$

Note that since $\cos 2 \theta = \dfrac{3}{5}$ , then $\cos^2 \theta = \dfrac{4}{5}$ and $\sin^2 \theta = \dfrac{1}{5}$ , and hence,

$\frac{4}{5} a^2 + \frac{1}{5} b^2 = \dfrac{37^2}{4}$

$\frac{1}{5} a^2 + \frac{4}{5} b^2 = \dfrac{26^2}{4}$

from which, and using Cramer's rule,

$a^2 = \dfrac{ \frac{1}{20} ( 4 (37)^2 - (26)^2 ) } { \frac{16}{25} - \frac{1}{25} } = 400$ , hence, $a=\boxed{20}$

Let us consider the standard ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ , with $a>b$ . Tilting the ellipse about the origin so that the semi-major axis makes an angle $\theta$ with the $x$ -axis, the equation becomes $\dfrac {(x\cos \theta + y\sin \theta)^2}{a^2} + \dfrac {(x\sin \theta - y\cos \theta)^2}{b^2} = 1$ . Shifting the center of the ellipse to $(x_c, y_c)$ , the equation becomes $\dfrac {\left((x-x_c)\cos \theta + (y-y_c) \sin \theta\right))^2}{a^2} + \dfrac {\left((x-x_c) \sin \theta - (y-y_c) \cos \theta\right))^2}{b^2} = 1$ . For $\theta = \frac 12 \cos^{-1} \frac 35$ $= \tan^{-1} \frac 12$ $= \sin^{-1} \frac 1{\sqrt 5}$ $= \cos^{-1} \frac 2{\sqrt 5}$ and $(x_c, y_c) = \left(\frac {37}2, 13\right)$ , the equation of the ellipse of this problem is:

$\begin{aligned} \frac {(2x+y-50)^2}{a^2} + \frac {(x-2y+7.5)^2}{b^2} & = 5 \quad ...(1) & \small \color{#3D99F6} \text{Differentiate w.r.t. }x \\ \frac {(2x+y-50)\left(2+\frac {dy}{dx}\right)}{a^2} + \frac {(x-2y+7.5)\left(1-2\frac {dy}{dx}\right)}{b^2} & = 0 \quad ...(2a) & \small \color{#3D99F6} \text{Differentiate w.r.t. }y \\ \frac {(2x+y-50)\left(2\frac {dx}{dy}+1 \right)}{a^2} + \frac {(x-2y+7.5)\left(\frac {dx}{dy} - 2\right)}{b^2} & = 0 \quad ...(2b) \end{aligned}$

Let the point where the ellipse touches the $x$ -axis be $(x_0, 0)$ . Note that $\dfrac {dy}{dx}= 0$ at the point. Then

$\begin{cases} \dfrac {(2x_0-50)^2}{a^2} + \dfrac {(x_0+7.5)^2}{b^2} = 5 & ...(1a) \\ \dfrac {2(2x_0-50)}{a^2} + \dfrac {x_0+7.5}{b^2} = 0 & ...(2aa) \end{cases}$

$\begin{aligned} (1a) - (x_0+7.5)(2aa): \ \ \frac {2x_0-50}{a^2} & = - \frac 1{13} & \small \color{#3D99F6} \implies x_0 = \frac {650-a^2}{26} \end{aligned}$

$\begin{aligned} (2aa): \ \ -\frac 2{13} + \frac {845-a^2}{26b^2} & = 0 & \color{#3D99F6} \implies b^2 = \frac {845-a^2}4 \ \ ...(3a) \end{aligned}$

Similarly, let the point where the ellipse touches the $y$ -axis be $(0, y_0)$ . Note that $\dfrac {dx}{dy}= 0$ at the point. Then

$\begin{cases} \dfrac {(y_0-50)^2}{a^2} + \dfrac {(7.5-2y_0)^2}{b^2} = 5 & ...(1b) \\ \dfrac {y_0-50}{a^2} - \dfrac {2(7.5-2y_0)}{b^2} = 0 & ...(2bb) \end{cases}$

$\begin{aligned} 2(1b) + (7.5-2y_0)(2bb): \ \ \frac {y_0-50}{a^2} & = - \frac 4{37} & \small \color{#3D99F6} \implies y_0 = \frac {1850-4a^2}{37} \end{aligned}$

$\begin{aligned} (2bb): \ \ -\frac 4{37} + \frac {6845-16a^2}{37b^2} & = 0 & \color{#3D99F6} \implies b^2 = \frac {6845-16a^2}4 \ \ ...(3b) \end{aligned}$

From $(3a) = (3b): \ \ 845 - a^2 = 6845 - 16a^2 \implies$ the semi-major axis $a = \boxed{20}$ .