4. For the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, the left focus is $F$, and the intersection point of the line $y=\sqrt{3} x$ with the ellipse in the first quadrant is $M$. A tangent line to the ellipse at $M$ intersects the $x$-axis at point $N$. Then the area of $\triangle M N F$ is ( ).
(A) $\sqrt{3}+\frac{\sqrt{15}}{5}$
(B) $\sqrt{3}+\frac{2 \sqrt{15}}{5}$
(C) $2 \sqrt{3}+\frac{\sqrt{15}}{5}$
(D) $2 \sqrt{3}+\frac{2 \sqrt{15}}{5}$
4.C.

As shown in Figure 4, from the ellipse equation $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, we get $a^{2}=4, b^{2}=3, c^{2}=1$.
Therefore, $F(-1,0)$.
Let $M\left(x_{0}, y_{0}\right)$, it is easy to know that the tangent line equation through point $M$ is
$$
\frac{x_{0} x}{4}+\frac{y_{0} y}{3}=1 .
$$

Let $y=0$, solving we get
$$
x_{N}=\frac{4}{x_{0}} \text {. }
$$

Thus, $|F N|=1+\frac{4}{x_{0}}$.
Therefore, $S_{\triangle M N F}=\frac{1}{2}\left(1+\frac{4}{x_{0}}\right) y_{0}=\frac{1}{2} y_{0}+\frac{2 y_{0}}{x_{0}}$. Let $x_{0}=2 \cos \theta, y_{0}=\sqrt{3} \sin \theta$.

Since $\frac{y_{0}}{x_{0}}=\frac{\sqrt{3}}{2} \tan \theta=\sqrt{3}$, we have $\tan \theta=2$.
Clearly, $\theta \in\left(0, \frac{\pi}{2}\right)$, then
$$
\sin \theta=\frac{2 \sqrt{5}}{5}, y_{0}=\sqrt{3} \sin \theta=\frac{2 \sqrt{15}}{5} \text {. }
$$

Therefore, $S_{\triangle M N F}=\frac{\sqrt{15}}{5}+2 \sqrt{3}$.