5. Given two ellipses $C_{1}$ and $C_{2}$, and $C_{1}$ is inside $C_{2}$. Let the equation of the inner ellipse $C_{1}$ be $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>$ $n>0)$, and the equation of the outer ellipse $C_{2}$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$. Draw a tangent line to $C_{1}$ at any point $M$ on $C_{1}$, which intersects ellipse $C_{2}$ at points $P$ and $Q$. Draw tangents to $C_{2}$ at points $P$ and $Q$. Then the locus of the intersection point $R$ of these two tangents is $(\quad)$.
(A) $\frac{x^{2}}{m^{2}} a^{2}+\frac{y^{2}}{n^{2} b^{2}}=1$
(B) $\frac{x^{2}}{\frac{a^{2}}{m^{2}}}+\frac{y^{2}}{\frac{b^{2}}{n^{2}}}=1$
(D) $\frac{x^{2}}{\frac{a^{4}}{m^{2}}}+\frac{y^{2}}{\frac{b^{4}}{n^{2}}}=1$
5. D.

Let $M\left(x_{0}, y_{0}\right)$, then
$$
l_{P Q}: \frac{x_{0} x}{m^{2}}+\frac{y_{0} y}{n^{2}}=1 \text {. }
$$

Suppose $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right), R\left(x_{R}, y_{R}\right)$. The equations of the tangents at points $P 、 Q$ are respectively
$$
\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1, \frac{x_{2} x}{a^{2}}+\frac{y_{2} y}{b^{2}}=1 \text {. }
$$

Since the intersection point of these two tangents is $R$, we have
$$
\frac{x_{1} x_{R}}{a^{2}}+\frac{y_{1} y_{R}}{b^{2}}=1, \frac{x_{2} x_{R}}{a^{2}}+\frac{y_{2} y_{R}}{b^{2}}=1 \text {. }
$$

Therefore, $l_{P Q}$ can also be expressed as $\frac{x_{R} x}{a^{2}}+\frac{y_{R} y}{b^{2}}=1$.
From equation (1), we know $\frac{x_{0}}{m^{2}}=\frac{x_{R}}{a^{2}}, \frac{y_{0}}{n^{2}}=\frac{y_{R}}{b^{2}}$.
Since $\frac{x_{0}^{2}}{m^{2}}+\frac{y_{0}^{2}}{n^{2}}=1$, eliminating $x_{0} 、 y_{0}$ and simplifying, we get
$$
\frac{x_{R}^{2}}{\frac{a^{4}}{m^{2}}}+\frac{y_{R}^{2}}{\frac{b^{4}}{n^{2}}}=1 .
$$