From Semi-Axes Sum to Conjugate Semi-Diameters

Re-arranging the given vector equation using the unit vector $\mathbf{u} = [ \cos t , \sin t ]^T$ , we have,

$\mathbf{r} - \mathbf{C} = \mathbf{F} \mathbf{u}$

where the columns of matrix $\mathbf{F}$ are the vectors $\mathbf{f_1}$ and $\mathbf{f_2}$ .

Now, since $\mathbf{u}^T \mathbf{u} = 1$ , the above vector equation becomes the algebraic equation,

$(\mathbf{r} - \mathbf{C})^T \mathbf{F}^{-T} \mathbf{F}^{-1} (\mathbf{r} - \mathbf{C}) = 1$

It follows that the eigenvalues of the inverse of the matrix $\mathbf{F}^{-T} \mathbf{F}^{-1}$ , which is the matrix $\mathbf{F} \mathbf{F}^T$ are $a^2$ and $b^2$ .

$\mathbf{F} \mathbf{F}^T = \begin{bmatrix} 7 && 3 \\ 10 && n \end{bmatrix} \begin{bmatrix} 7 && 10 \\ 3 && n \end{bmatrix} = \begin{bmatrix} 58 && 70+3n \\ 70+3n && 100+n^2 \end{bmatrix}$

The characteristic equation of this matrix is $(\lambda - 58)(\lambda - (100+n^2) ) - (70+3n)^2 = 0$

Since the two roots of this equation are $a^2$ and $b^2$ , then

$a^2 + b^2 = 158 + n^2$

and

$a^2 b^2 = 58(100+n^2) - (70+3n)^2 = 49 n^2 - 420 n + 900$

In addition, we are given that $a + b = 25$ , hence,

$(a + b)^2 = a^2 + b^2 + 2 a b$

Therefore,

$625 = 158 + n^2 + 2 \sqrt{ 49 n^2 - 420 n + 900 }$

Re-arranging this, we get

$4 ( 49 n^2 - 420 n + 900) = ( 467 - n^2 )^2$

A quartic equation with four real roots, two of which are valid and these are $-14.3541565$ , and $17$ and two extraneous solutions which are $-31$ , and $28.3541565$ . Since we are given that $n$ is an integer, then the only solution is $\boxed{17}$

We Bengalees have an old proverb : frying a fish in it's own fat . Doing the same here!

Following you, the semi-axes are given by

$a, b=\dfrac {1}{\sqrt 2}\sqrt {f_{1x}^2+f_{1y}^2+f_{2x}^2+f_{2y}^2\pm \sqrt {\left ((f_{1x}-f_{2y})^2+(f_{1y}+f_{2x})^2\right )\left ((f_{1x}+f_{2y})^2+(f_{1y}-f_{2x})^2\right )}}$

$=\dfrac 12 \left (\sqrt {n^2+14n+98}\pm \sqrt {n^2-14n+218}\right )$

$\implies a+b=\sqrt {n^2+14n+98}=25$ (given)

$\implies n^2+14n-527=0$

The positive solution to this equation is $n=\boxed {17}$ .