10. Through the point \( M(3,2) \) inside the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\), draw a line \( A B \) intersecting the ellipse at points \( A \) and \( B \), and draw a line \( C D \) intersecting the ellipse at points \( C \) and \( D \). Draw tangents to the ellipse at \( A \) and \( B \) intersecting at \( P \), and draw tangents to the ellipse at \( C \) and \( D \) intersecting at \( Q \). Find the equation of the line connecting \( P \) and \( Q \).
10. Solution: The tangent line equations through points $A, B, C, D$ are
$$
l_{P A}: \frac{x_{A} x}{25}+\frac{y_{A} y}{9}=1, \quad l_{P B}: \frac{x_{B} x}{25}+\frac{y_{B} y}{9}=1, \quad l_{\mathrm{QC}}: \frac{x_{\mathrm{C}} x}{25}+\frac{y_{C} y}{9}=1,
$$
$$
I_{Q D}: \frac{x_{D} x}{25}+\frac{y_{D} y}{9}=1,
$$
Since point $P\left(x_{P}, y_{P}\right)$ lies on $P A$ and $P B$, we have $\frac{x_{A} x_{P}}{25}+\frac{y_{A} y_{P}}{9}=1, \frac{x_{B} x_{P}}{25}+\frac{y_{B} y_{P}}{9}=1$.
This indicates that $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$ lie on the line $\frac{x_{P} x}{25}+\frac{y_{P} y}{9}=1$; similarly, the line equation for $C D$ is: $\frac{x_{Q} x}{25}+\frac{y_{Q} y}{9}=1$;
Since $A B$ and $C D$ intersect at point $M(3,2)$, we have $\frac{3 x_{P}}{25}+\frac{2 y_{P}}{9}=1, \frac{3 x_{Q}}{25}+\frac{2 y_{Q}}{9}=1$,
Therefore, the line equation for $P$ and $Q$ is $\frac{3 x}{25}+\frac{2 y}{9}=1$.