Don't wanna bash!

Equation of first ellipse (assumed WLOG in standard position):

$\dfrac{x^2}{ a^2} + \dfrac{y^2}{b^2} = 1$

Equation of second ellipse (before rotation)

$\dfrac{x^2} {a'^2} + \dfrac{y^2}{b'^2} = 1$

After rotation by an angle $\theta$ the second ellipse has the equation:

$r^T R D R^T r = 1$

with

$D = \text{diag} \{ \dfrac{1}{a'^2} , \dfrac{1}{b'^2} \}$

and

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

Now let $c = \cos \theta$ and $s = \sin \theta$ .

Performing the multiplication $R D R^T$ we find that

$R D R^T = \begin{bmatrix} \dfrac{c^2}{ a'^2} + \dfrac{s^2}{ b'^2} && (\dfrac{1}{a'^2} - \dfrac{1}{b'^2}) s c \\ (\dfrac{1}{a'^2} - \dfrac{1}{b'^2}) s c && \dfrac{c^2}{ b'^2} + \dfrac{s^2}{ a'^2} \end{bmatrix}$

Focii of first ellipse are at $( a u, 0)$ and $(-au, 0)$ .

Focii of second ellipse are at $(a' v, 0)$ and $(-a' v, 0)$ rotated by $\theta$ , i.e. $(a'v c, a'v s)$ and $(-a' v c, -a' v s )$

Substituting the coordinates of one of the focii of the first ellipse into the equation of the second ellipse,

$( \dfrac{c^2}{ a'^2} + \dfrac{s^2}{ b'^2}) (a^2 u^2) = 1$

Similarly, substituting the coordinates of one of the focii of the second ellipse into the equation of the first ellipse,

$(a'^2 v^2) ( \dfrac{c^2}{ a^2} + \dfrac{s^2}{ b^2} )= 1$

Now, from the definition of eccentricity, we have $u = \sqrt{1 - (b/a)^2}$ ,therefore,

$b^2 = a^2 (1 - u^2)$ , and similarly, $b'^2 = a'^2 (1 - v^2)$

Substituting this, we arrive at,

$\dfrac{a^2 u^2}{ a'^2} ( c^2 + \dfrac{s^2}{ (1 - v^2) }) = 1$

$\dfrac{a'^2 v^2}{ a^2} ( c^2 + \dfrac{s^2}{ (1 - u^2)} ) = 1$

Multiplying the two equations:

$(uv)^2 ( c^2 + \dfrac{s^2}{ (1 - v^2)} )(c^2 + \dfrac{ s^2 }{ (1 - u^2)} ) = 1$

Which can be simplified as follows:

$(uv)^2 ( c^2 (1 - v^2) + s^2 ) (c^2 (1 - u^2) + s^2 ) = (1 - u^2 )(1 - v^2)$

$(uv)^2 ( 1 - c^2 v^2) (1 - c^2 u^2 ) = (1 - (u^2 + v^2) + (uv)^2 )$

$(uv)^2 ( 1 - c^2 (u^2 + v^2) + c^4 (uv)^2 ) = 1 - (u^2 + v^2) + (uv)^2$

Now, let $e = uv$ and $S = u^2 + v^2$ , then

$e^2 (1 - c^2 S + c^4 e^2 ) = 1 - S + e^2$

which implies

$e^2 ( - c^2 S + c^4 e^2 ) = 1 - S$

Hence,

$S = \dfrac{(1 - e^4 c^4)}{ (1 - e^2 c^2) }= 1 + e^2 c^2$

Finally, substituting $e = \dfrac{3}{4}$ and $c = \dfrac{\sqrt{8}}{3}$ we get

$S = 1.5$