Since $\sqrt{2} + \sqrt{3} > \pi$, so $0 < \frac{\pi}{2} - \sqrt{2} < \sqrt{3} - \frac{\pi}{2} < \frac{\pi}{2}$ and
$$\cos(\frac{\pi}{2} - \sqrt{2}) > \cos(\sqrt{3} - \frac{\pi}{2}), \text{ i.e. } \sin\sqrt{2} > \sin\sqrt{3}.$$

Since
$$(\sin\sqrt{2} - \sin\sqrt{3}) - (\cos\sqrt{2} - \cos\sqrt{3}) = 2\sqrt{2}\sin\frac{\sqrt{2}-\sqrt{3}}{2}\sin\left(\frac{\sqrt{2}+\sqrt{3}}{2} + \frac{\pi}{4}\right) \quad (*)$$
and
$$-\frac{\pi}{2} < \frac{\sqrt{2}-\sqrt{3}}{2} < 0,$$
we get
$$\sin\frac{\sqrt{2}-\sqrt{3}}{2} < 0, \quad \frac{\pi}{2} < \frac{\sqrt{2}+\sqrt{3}}{2} < \frac{3\pi}{4},$$
$$\frac{3\pi}{4} < \frac{\sqrt{2}+\sqrt{3}}{2} + \frac{\pi}{4} < \pi,$$
$$\sin\left(\frac{\sqrt{2}+\sqrt{3}}{2} + \frac{\pi}{4}\right) > 0,$$

so the expression $(*)$ is less than $0$.
That is $\sin\sqrt{2} - \sin\sqrt{3} < \cos\sqrt{3} - \cos\sqrt{2}$, therefore the curve is an ellipse with foci on the y-axes. Answer: C.