Square in a Quarter Ellipse

Consider the general case, where the ellipse has semimajor and semiminor axes $a,b$ . Say the two points on the curve are $P(a \cos u,B \sin u),\;\; Q(a \cos v,b \sin v)$

where $u>v$ (so $P$ is the point closer to the $y$ -axis), and define $R$ and $S$ as below:

Since $PQRS$ is a square, the difference in the $y$ -coordinates of points $Q$ and $R$ is the same as the difference in the $x$ -coordinates of $Q$ and $P$ , etc; so we have $R(a \cos v -b \sin u+b \sin v, b \sin v-a \cos v+a \cos u)$ and $S(a \cos u -b \sin u+b \sin v, b \sin u-a \cos v+a \cos u)$

Since $R$ is on the $x$ -axis and $S$ on the $y$ -axis, we have $\begin{aligned} b \sin v-a \cos v+a \cos u &=0 \\ a \cos u -b \sin u+b \sin v &=0 \end{aligned}$ from which we see that $a\cos v=b\sin u$ so we can write $v$ in terms of $u$ ; in particular $\sin v=\sqrt{1-\frac{b^2 \sin^2 u}{a^2}}$

Substituting into the second equation above, we get $\begin{aligned} a \cos u -b\sin u+b \sqrt{1-\frac{b^2 \sin^2 u}{a^2}} &=0 \end{aligned}$

We could solve this numerically for $u$ here; but amazingly, it is solvable analytically. If we let $x=\sin u$ (so that $\cos u=\sqrt{1-x^2}$ ) and clear square roots, we find that $x$ is a root of $\left(a^4 - a^2 b^2 + b^4\right) \left(a^4 + 3 a^2 b^2 + b^4\right)x^4-2a^2\left(a^6+b^6\right)x^2 + a^4 \left(a^2 - b^2\right)^2=0$

This looks bad, but is just a biquadratic in $x$ (that is, a quadratic in $x^2$ ). In this case, the $a=12$ and $b=7$ and the polynomial is $712468705 x^4- 893846304 x^2+187142400=0$

The root we want is $x= 12 \sqrt{\frac{193}{44305} + \frac{14112}{44305 \sqrt{16081}}}$

From this we can calculate $u$ , $v$ , the coordinates of the vertices of the square, and all the parameters we need for the solution; we find $s=5.83229,\;\;C_x=C_y=3.48066,\;\;\theta=12.45708^\circ$ for a total of $\boxed{25.25764\ldots}$ .