Orthoptic Sharing

If a and b be the major and minor semi-axes of the ellipse, and c and d be those of the hyperbola, then $a^2+b^2$ = $c^2-d^2$ . Also, let a=2p and c=3p, where p is a natural number. Using the given points on the curves, we have $45p^4-1738p^2+15325=0$ . From this we get p=5. Hence k = $a^2+b^2$ =125

Let $p_1$ and $q_1$ be $a_e^2$ and $b_e^2$ of the ellipse, and $p_2$ and $q_2$ be $a_h^2$ and $b_h^2$ of the hyperbola.

By the properties of orthoptic circles,

$p_1 + q_1 = p_2 - q_2$

Since $(6, 4)$ is on the ellipse,

$\frac{36}{p_1} + \frac{16}{q_1} = 1$

Since $(17, \frac{16}{3})$ is on the ellipse,

$\frac{289}{p_2} - \frac{256}{9q_2} = 1$

and since the semi-major axes have a $\frac{2}{3}$ ratio,

$p_1 = \frac{4}{9}p_2$

These equations combine to:

$p_1(p_1 - 100)(45p_1 - 2452) = 0$

which has one positive integer solution $p_1 = 100$ .

Since $p_1 = 100$ and $\frac{36}{p_1} + \frac{16}{q_1} = 1$ , then $q_1 = 25$ , and since $r^2$ of the orthoptic circle is $p_1 + q_1 = 100 + 25 = 125$ , its area is $\pi r^2 = 125 \pi$ , so that $k = \boxed{125}$ .