Find the equation of the ellipse:
The equation of a standard ellipse centered at the origin is $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ From the diagram we can infer that $c=20$, which is $c^2=400=a^2-b^2$. The coordinates of $A$ should satisfy the equation of the ellipse, so $$\dfrac{169}{a^2}+\dfrac{432}{b^2}=1\iff169b^2+432a^2=a^2b^2.$$ Now we are left with finding $a^2$ and $b^2$ from $$\begin{cases}a^2-b^2=400\Rightarrow a^2=400+b^2\\169b^2+432a^2=a^2b^2\end{cases}$$ We get $169b^2+432(400+b^2)=b^2(400+b^2)$ which is biquadratic for $b$. Is there a way to find $a^2$ and $b^2$ without such gross calculations?
The ellipse is the locus of constant foci distance sum.
As you further already said: $c^2=a^2-b^2$, thus by symmetry it becomes clear that this sum $s$ is related by $(s/2)^2 = c^2+b^2=a^2$, i.e. $s=2a$.
Thus $$a=\frac{dist(F_1,A)+dist(A,F_2)}2$$
--- rk