From conjugate semi-diameters to semi-axes

Translating the center to the point $C$ , we can write down the equation of the ellipse in the parametric form as: $x(t) = 3 \cos(t), y(t) = \cos(t) + 2 \sin(t).$ Hence, in polar coordinates, the length of the radius vector is given by $r(t) = \sqrt{x^2(t)+ y(t)^2}= \sqrt{7+ \sqrt{13}\cos(2t+ \alpha)},$ for some angle $\alpha$ . The length of the semi-major and semi-monor axes are given by the maximum and minimum values of $r(t)$ respectively. Hence, their sum is $\sqrt{7+ \sqrt{13}} + \sqrt{7- \sqrt{13}} \approx 5.099.$

We can ignore $\mathbf{c}$ , since it just shifts the centre of the ellipse. Thus we have $x \; =\; 3\cos t \hspace{2cm} y \; = \; \cos t + 2\sin t$ and hence $36 \; = \; 4x^2 + (3y - x)^2 \; = \; 5x^2 - 6xy + 9y^2 \; = \; \big(\,x\;\;y\,\big)\left(\begin{array}{cc} 5 & -3 \\ -3 & 9 \end{array}\right)\left(\begin{array}{c} x \\ y \end{array}\right) \; = \; \big(\,x\;\;y\,\big)P^T\left(\begin{array}{cc} u & 0 \\ 0 & v \end{array}\right)P\left(\begin{array}{c} x \\ y \end{array}\right)$ for some orthogonal matrix $P$ , where $u,v$ are the eigenvalues of the matrix $\left(\begin{array}{cc} 5 & -3 \\ -3 & 9 \end{array}\right)$ Thus the semimajor and semiminor axes $a,b$ are $\tfrac{6}{\sqrt{u}}$ and $\tfrac{6}{\sqrt{v}}$ , and hence we want to find $a + b \; =\; \frac{6}{\sqrt{u}} + \frac{6}{\sqrt{v}} \; = \; \frac{6(\sqrt{u}+\sqrt{v})}{\sqrt{uv}}$ noting that $(a+b)^2 \; = \; \frac{36}{uv}\big(u + v + 2\sqrt{uv}\big)$ Now $u,v$ are the roots of the characteristic polynomial $(t-5)(t-9)-9 \; = \; t^2 - 14t + 36$ so that $u+v=14,uv=36$ and hence $(a+b)^2 = 14 +2\times6 =26$ making the answer $\boxed{\sqrt{26}}$ .

For an ellipse $\overrightarrow{p}(t) = \overrightarrow{f_0} + \overrightarrow{f_1} \cos t + \overrightarrow{f_2} \sin t$ , the four vertices are $\overrightarrow{p}(t_0)$ , $\overrightarrow{p}(t_0 \pm \frac{\pi}{2})$ , and $\overrightarrow{p}(t_0 + \pi)$ where $t_0$ is defined by $\cot (2t_0) = \frac{\overrightarrow{f_1}^2 - \overrightarrow{f_2}^2}{2\overrightarrow{f_1} \cdot \overrightarrow{f_2}}$ (see here ).

Using $\overrightarrow{f_1} = (3, 1)$ and $\overrightarrow{f_2} = (0, 2)$ , $\cot (2t_0) = \frac{(3^2 + 1^2) - (2^2 - 0^2)}{2(3 \cdot 0 + 1 \cdot 2)} = \frac{3}{2}$ or $\tan (2t_0) = \frac{2}{3}$ , and using some trig identities:

$\tan t_0 = \frac{-3 + \sqrt{13}}{2}$

$\sin t_0 = \sqrt{\frac{13-3\sqrt{13}}{26}}$

$\cos t_0 = \sqrt{\frac{13+3\sqrt{13}}{26}}$

Since the endpoints of the major axis are defined by $\overrightarrow{p}(t_0)$ and $\overrightarrow{p}(t_0 + \pi)$ :

semi-major axis $a$

$= \frac{1}{2}|\overrightarrow{p}(t_0) - \overrightarrow{p}(t_0 + \pi)|$

$= \frac{1}{2}|(\overrightarrow{f_0} + (3, 1) \cos t_0 + (0, 2) \sin t_0) - (\overrightarrow{f_0} + (3, 1) \cos (t_0 + \pi) + (0, 2) \sin (t_0 + \pi))|$

$= |(3, 1) \cos t_0 + (0, 2) \sin t_0|$

$= |(3, 1) \sqrt{\frac{13+3\sqrt{13}}{26}} + (0, 2) \sqrt{\frac{13-3\sqrt{13}}{26}}|$

$= \sqrt{(3 \cdot \sqrt{\frac{13+3\sqrt{13}}{26}})^2 + (1 \cdot \sqrt{\frac{13+3\sqrt{13}}{26}} + 2 \cdot \sqrt{\frac{13-3\sqrt{13}}{26}})^2}$

$= \sqrt{7 + \sqrt{13}}$

Since the endpoints of the minor axis are defined by $\overrightarrow{p}(t_0 \pm \frac{\pi}{2})$ :

semi-minor axis $b$

$= \frac{1}{2}|\overrightarrow{p}(t_0 - \frac{\pi}{2}) - \overrightarrow{p}(t_0 + \frac{\pi}{2})|$

$= \frac{1}{2}|(\overrightarrow{f_0} + (3, 1) \cos (t_0 - \frac{\pi}{2}) + (0, 2) \sin (t_0 - \frac{\pi}{2})) - (\overrightarrow{f_0} + (3, 1) \cos (t_0 + \frac{\pi}{2}) + (0, 2) \sin (t_0 + \frac{\pi}{2}))|$

$= |(3, 1) \sin t_0 - (0, 2) \cos t_0|$

$= |(3, 1) \sqrt{\frac{13-3\sqrt{13}}{26}} + (0, 2) \sqrt{\frac{13+3\sqrt{13}}{26}}|$

$= \sqrt{(3 \cdot \sqrt{\frac{13-3\sqrt{13}}{26}})^2 + (1 \cdot \sqrt{\frac{13-3\sqrt{13}}{26}} + 2 \cdot \sqrt{\frac{13+3\sqrt{13}}{26}})^2}$

$= \sqrt{7 - \sqrt{13}}$

Therefore, the sum of the semi-axes lengths is $a + b = \sqrt{7 + \sqrt{13}} + \sqrt{7 - \sqrt{13}} \approx \boxed{5.099}$ .

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Using the same method above with $\overrightarrow{f_1} = (f_{1x}, f_{1y})$ and $\overrightarrow{f_2} = (f_{2x}, f_{2y})$ , the semi-axes lengths are:

$\sqrt{\frac{1}{2}(f_{1x}^2 + f_{1y}^2 + f_{2x}^2 + f_{2y}^2) \pm \frac{1}{2}\sqrt{((f_{1x} - f_{2y})^2 + (f_{1y} + f_{2x})^2)((f_{1x} + f_{2y})^2 + (f_{1y} - f_{2x})^2)}}$

and the sum of the semi-axes lengths are:

$\max(\sqrt{(f_{1x} + f_{2y})^2 + (f_{1y} - f_{2x})^2}, \sqrt{(f_{1x} - f_{2y})^2 + (f_{1y} + f_{2x})^2})$