9. (16 points) Given the parabola $y=x^{2}$ and the ellipse $(x-3)^{2}+4 y^{2}=1$, the focus of the parabola is $F$, and $A$ is a point on the ellipse. The line $F A$ intersects the parabola at points $B$ and $C$. Tangents to the parabola are drawn through points $B$ and $C$. Find the locus of the intersection of the two tangents.
9. Let the intersection point of the two tangents be \( M\left(x_{0}, y_{0}\right) \), and the coordinates of points \( B \) and \( C \) be \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \), respectively. It is easy to know that the equation of the tangent at point \( B \) is \( \frac{y+y_{1}}{2}=x_{1} x \), and the equation of the tangent at point \( C \) is \( \frac{y+y_{2}}{2}=x_{2} x \). Substituting the point \( M\left(x_{0}, y_{0}\right) \) into the above two tangent equations, we find that points \( B \) and \( C \) both satisfy the equation \( \frac{y+y_{0}}{2}=x_{0} x \), i.e.,
$$
y=2 x_{0} x-y_{0},
$$

which is the equation of line \( F A \).
On the other hand, the coordinates of point \( A \) can be expressed as \( \left(3+\cos \theta, \frac{1}{2} \sin \theta\right) (0 \leqslant \theta \leqslant 2 \pi) \), and it is easy to know that point \( F\left(0, \frac{1}{4}\right) \). Therefore, the equation of line \( F A \) can be expressed as
$$
y=\frac{\frac{1}{2} \sin \theta-\frac{1}{4}}{3+\cos \theta} x+\frac{1}{4} .
$$

Comparing the two equations of line \( F A \), we get
$$
\left\{\begin{array}{l}
x_{0}=\frac{1}{4} \cdot \frac{\sin \theta-\frac{1}{2}}{3+\cos \theta}, \\
y_{0}=-\frac{1}{4} .
\end{array}\right.
$$

Let the function
$$
u=\frac{\sin \theta-\frac{1}{2}}{3+\cos \theta}
$$

then
$$
\sin \theta-u \cos \theta=3 u+\frac{1}{2},
$$

i.e., \( \sqrt{1+u^{2}} \sin (\theta+\varphi)=3 u+\frac{1}{2} \), where \( \sin \varphi=-\frac{u}{\sqrt{u^{2}+1}}, \cos \varphi=\frac{1}{\sqrt{u^{2}+1}} \). Thus, by the boundedness of trigonometric functions, we have
$$
\left|\frac{3 u+\frac{1}{2}}{\sqrt{u^{2}+1}}\right| \leqslant 1,
$$

Solving this, we get \( \frac{-3-\sqrt{33}}{16} \leqslant u \leqslant \frac{-3+\sqrt{33}}{16} \), and thus \( \frac{-3-\sqrt{33}}{64} \leqslant x_{0} \leqslant \frac{-3+\sqrt{33}}{64} \). Therefore, the trajectory of point \( M \) is the line segment
$$
y=-\frac{1}{4} \quad\left(\frac{-3-\sqrt{33}}{64} \leqslant x \leqslant \frac{-3+\sqrt{33}}{64}\right) .
$$