10. (20 points) Given that the ellipse $\Gamma_{1}$ and the hyperbola $\Gamma_{2}$, both centered at the origin, have a common focus $F(1,0)$, and both pass through the point $P\left(1, \frac{3}{2}\right)$.
(1) Find the standard equations of $\Gamma_{1}$ and $\Gamma_{2}$.
(2) A line passing through point $F$ intersects $\Gamma_{1}$ at points $A$ and $B$, and intersects $\Gamma_{2}$ at points $C$ and $D$. Does there exist such a line such that $|A B|=2|C D|$? Explain your reasoning.
10. (1) Let $\Gamma_{1}: \frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1, \Gamma_{2}: \frac{x^{2}}{a_{2}^{2}}-\frac{y^{2}}{b_{2}^{2}}=1$.

By symmetry, let the other focus be $F_{1}(-1,0)$, with semi-focal distance $c=1$.
Then $\left|P F_{1}\right|=\frac{5}{2},|P F|=\frac{3}{2}$.
Thus, $2 a_{1}=\left|P F_{1}\right|+|P F|=4$,
$2 a_{2}=\left|P F_{1}\right|-|P F|=1$.
Therefore, $a_{1}=2, b_{1}=\sqrt{3}, a_{2}=\frac{1}{2}, b_{2}=\frac{\sqrt{3}}{2}$.
Hence, $\Gamma_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1, \Gamma_{2}: 4 x^{2}-\frac{4 y^{2}}{3}=1$.
(2) Let the line be $l$. If line $l$ is horizontal, then $|A B|=4,|C D|=1$, which does not meet the condition.
Thus, let its equation be $x=m y+1$.
From $\left\{\begin{array}{l}x=m y+1, \\ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1,\end{array}\right.$,
$\left(3 m^{2}+4\right) y^{2}+6 m y-9=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$\left|y_{1}-y_{2}\right|=\frac{12 \sqrt{m^{2}+1}}{3 m^{2}+4}$.
From $\left\{\begin{array}{l}x=m y+1, \\ 4 x^{2}-\frac{4 y^{2}}{3}=1,\end{array}\right.$ we get
$\left(12 m^{2}-4\right) y^{2}+24 m y+9=0$.
Thus, $m^{2} \neq \frac{1}{3}$,
Discriminant $\Delta^{\prime}=12^{2}\left(m^{2}+1\right)>0$.
Let $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$. Then
$\left|y_{3}-y_{4}\right|=\frac{3 \sqrt{m^{2}+1}}{\left|3 m^{2}-1\right|}$.
Thus, $\frac{|A B|}{|C D|}=\frac{\left|y_{1}-y_{2}\right|}{\left|y_{3}-y_{4}\right|}=4 \times \frac{\left|3 m^{2}-1\right|}{3 m^{2}+4}$.
If $|A B|=2|C D|$, then
$4 \times \frac{\left|3 m^{2}-1\right|}{3 m^{2}+4}=2 \Rightarrow m^{2}=2$.
Therefore, the equation of the required line is $x \pm \sqrt{2} y-1=0$.