2. The two roots of the equation
$$
\frac{x^{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right) x+\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=0
$$

are the eccentricities of two conic sections, which are ( ).
(A) ellipse and hyperbola
(B) hyperbola and parabola
(C) parabola and ellipse
(D) ellipse and ellipse
2.C.

Let $e_{1}$ and $e_{2}$ represent these two eccentricities. Then, from
$$
e_{1} e_{2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}<1
$$

we know that one of them must be an ellipse. Furthermore, from
$$
\begin{array}{l}
\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \\
=\frac{2(\sqrt{2}+\sqrt{3})}{(\sqrt{2}+\sqrt{3})^{2}-5}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{6}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}
\end{array}
$$

we know that 1 is a root of the equation, hence one of them is a parabola.