2. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have one focus at $F$, and point $P$ be on the $y$-axis. The line $P F$ intersects the ellipse at points $M$ and $N$, with $P M=\lambda_{1} M F$ and $P N=\lambda_{2} N F$. Then the real number $\lambda_{1}+\lambda_{2}=(\quad)$.
(A) $-\frac{2 b^{2}}{a^{2}}$
(B) $-\frac{b^{2}}{a^{2}}$
(C) $-\frac{2 a^{2}}{b^{2}}$
(D) $-\frac{a^{2}}{b^{2}}$
2.C.

Let's assume $F(c, 0)$.
Let $P(0, p)$ and $M(x, y)$. Suppose $P M=\lambda M F$, i.e., $(x, y-p)=\lambda(c-x,-y)$.
Solving this, we get $x=\frac{\lambda c}{1+\lambda}, y=\frac{p}{1+\lambda}$.
Substituting into the ellipse equation and simplifying, we get
$$
b^{4} \lambda^{2}+2 a^{2} b^{2} \lambda+a^{2} b^{2}-a^{2} p^{2}=0 .
$$

Thus, $\lambda_{1}+\lambda_{2}=-\frac{2 a^{2}}{b^{2}}$.