Elliptic Mania!

The whole diagram is symmetrical in a $45°$ line, so we can rotate the whole ellipse and place the bottom right point at $(0, 0)$ for a vertical ellipse with a center of $(b, 0)$ , as shown below.

The $j$ and $2j$ sides are hypotenuses of isosceles right triangles, and so we can deduce that the other points on the ellipse are $(\sqrt{2}j, \pm \sqrt{2}j)$ and $(\frac{3\sqrt{2}}{2}j, \pm \frac{\sqrt{2}}{2}j)$ .

Since the equation of a vertical ellipse with center $(b, 0)$ is $\frac{(x - b)^2}{b^2} + \frac{y^2}{a^2} = 1$ , we have $\frac{(\sqrt{2}j - b)^2}{b^2} + \frac{(\pm \sqrt{2}j)^2}{a^2} = 1$ and $\frac{(\frac{3\sqrt{2}}{2}j - b)^2}{b^2} + \frac{(\pm \frac{\sqrt{2}}{2}j)^2}{a^2} = 1$ .

These two equations and the given equation $a^2 + b^2 = 19200$ solve to $a = \pm 20 \sqrt{30}$ , $b = 60 \sqrt{2}$ , and $j = \boxed{75}$ for $j > 0$ .

If the area $A$ of the ellipse above

(1) $(j,j): j^2a + j^2b + j^2c + jd + je = 0$

(2) $(0,-j): jc - e = 0 \implies \boxed{c = \dfrac{e}{j}}$

(3) $2j,j): 4j^2a + 2j^2b + j^2c + 2jd + je = 0$

(4) $(2j,-j): 4j^2a - 2j^2b + j^2c + 2jd - je = 0$

Subtracting (4) from (3) $\implies 2jb + e = 0 \implies \boxed{b = -\dfrac{e}{2j}}$

Replacing $c = \dfrac{e}{j}$ and $b = -\dfrac{e}{2j}$ into equations (1) and (3) $\implies$

$2ja + 2d = -3e$

$4ja + 2d = -e$

$\implies \boxed{d = -\dfrac{5}{2}e}$ and $\boxed{a = \dfrac{e}{j}}$

$\implies \dfrac{1}{j}x^2 - \dfrac{1}{2j}xy + \dfrac{1}{j}y^2 - \dfrac{5}{2}x + y = 0$

$\implies 2x^2 - xy + 2y^2 - 5jx + 2jy = 0$

Let

$x = x'\cos(\theta) - y'\sin(\theta)$

$y = x'\sin(\theta) + y'\cos(\theta)$

$2(x'^2\cos^2(\theta) - \sin(2\theta)x'y' + y'^2\sin^2(\theta)) - ((\dfrac{x'^2 - y'^2}{2})\sin(2\theta) + \cos(2\theta)x'y')$ $+ 2(x'^2\cos^2(\theta)) + \sin(2\theta)x'y' + y'^2\sin^2(\theta)) - 5j(x'\cos(\theta) - y'\sin(\theta)) + 2j(x'\sin(\theta) + y'\cos(\theta)) = 0$

Setting $x'y'$ terms to zero $\implies \cos(2\theta) = 0 \implies \theta = \dfrac{\pi}{4}$

$\implies$

$x = \dfrac{1}{\sqrt{2}}(x' - y')$

$y = \dfrac{1}{\sqrt{2}}(x' + y')$

$\implies 3x'^2 + 5y'^2 - 3\sqrt{2}jx' + 7\sqrt{2}jy' = 0 \implies 3(x' - \dfrac{j}{\sqrt{2}})^2 + 5(y' + \dfrac{7}{5\sqrt{2}}j)^{2} = \dfrac{32j^{2}}{5}$

$\implies \dfrac{(x' - \dfrac{j}{\sqrt{2}})^2}{(\dfrac{32j^2}{15})} + \dfrac{(y' + \dfrac{7}{5\sqrt{2}}j)^2}{(\dfrac{32j^{2}}{25})}$

$\implies a = \dfrac{4\sqrt{2}j}{\sqrt{15}}$ and $b = \dfrac{4\sqrt{2}j}{5} \implies a^2 + b^2 = \dfrac{256}{75}j^{2} = 19200 \implies j = \boxed{75}$ .

Find the intersections of line y=x-b with ellipse as (0,-b) and {(2ba^2/(a^2+b^2)), b(a^2-b^2)/(a^2+b^2)}. The length of the chord between these two points is set equal to integer 2j.

Setting a^2+b^2=19200 and defining the i=(√2)b as the new constant , we get the following, Diophantine equation;

i^3=38400(i-j).

Solution is i=120 and j=75.

Answer=75.