15. (13 points) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$, the length of the minor axis is $2 \sqrt{3}$, and the eccentricity is $e_{1}$; the hyperbola $\frac{x^{2}}{m}-\frac{y^{2}}{n}=1(m 、 n>0)$ has asymptotes $y= \pm \sqrt{3} x$, and the eccentricity is $e_{2} ; e_{1} e_{2}=1$.
(1) Find the equation of the ellipse $E$;
(2) If $A$ is the right vertex of the ellipse $E$, $P\left(-1, \frac{3}{2}\right)$, the line $A P$ intersects the $y$-axis at point $H$, and another line through point $H$ intersects the ellipse $E$ at points $M$ and $N$. Let the area of $\triangle H M A$ be $S_{1}$, and the area of $\triangle P H N$ be $S_{2}$, and $S_{1}=6 S_{2}$, find the equation of the line $M N$.
15. (1) From the problem, we know
$2 b=2 \sqrt{3}, b=\sqrt{3}, \frac{n}{m}=3$,
$e_{2}=\sqrt{1+\frac{n}{m}}=2$.
Thus, $e_{1}=\frac{c}{a}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{e_{2}}=\frac{1}{2}$.
Therefore, $a=2$.
Hence, the standard equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) From (1), we know $P\left(-1, \frac{3}{2}\right)$ is on the ellipse $E$, and $A(2,0)$.
So, $l_{A P}: y=-\frac{1}{2}(x-2)$.
Thus, $H(0,1)$.
When $M N \perp x$-axis, it does not meet the problem's requirements.
When $M N$ is not perpendicular to the $x$-axis, let
$l_{M N}: y=k x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ 3 x^{2}+4 y^{2}=12,\end{array}\right.$ eliminating $y$, we get
$$
\left(4 k^{2}+3\right) x^{2}+8 k x-8=0 \text {. }
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{-8 k}{4 k^{2}+3}, \\
x_{1} x_{2}=\frac{-8}{4 k^{2}+3} .
\end{array}\right.
$$

From $S_{1}=6 S_{2} \Rightarrow|A H||M H|=6|N H||P H|$.
Since $|A H|=2|P H|$, we have
$$
|M H|=3|N H| \Rightarrow x_{1}=-3 x_{2} \text {. }
$$

Combining with the system of equations (1), we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
-2 x_{2}=\frac{-8 k}{4 k^{2}+3}, \\
-3 x_{2}^{2}=\frac{-8}{4 k^{2}+3}
\end{array}\right. \\
\Rightarrow k= \pm \frac{\sqrt{6}}{2} .
\end{array}
$$

Therefore, $l_{M N}: y= \pm \frac{\sqrt{6}}{2} x+1$.