11. A. As shown in Figure 5, the parabola $y=a x^{2}+b x(a>0)$ intersects the hyperbola $y=\frac{k}{x}$ at points $A$ and $B$. It is known that point $A(1,4)$, point $B$ is in the third quadrant, and the area of $\triangle A O B$ is 3 (where $O$ is the origin).
(1) Find the values of the real numbers $a$, $b$, and $k$;
(2) Draw a line $A C \parallel x$-axis through point $A$ on the parabola, intersecting the parabola at another point $C$. Find the coordinates of all points $E$ that satisfy $\triangle E O C \sim \triangle A O B$.
11. A. (1) Since point $A(1,4)$ lies on the hyperbola $y=\frac{k}{x}$, we have $k=4$.
Therefore, the function expression of the hyperbola is $y=\frac{4}{x}$.
Let point $B\left(t, \frac{4}{t}\right)(t>0)$, so,
$$
\left\{\begin{array}{l}
a + b = 4, \\
4a - 2b = -2
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b=3.
\end{array}\right.\right.
$$
(2) As shown in Figure 10.
Since $AC \parallel x$-axis, we have $C(-4,4)$.
Thus, $CO=4\sqrt{2}$.
Also, $BO=2\sqrt{2}$, so $\frac{CO}{BO}=2$.
Since $k_{OB}=1$ and $k_{OC}=-1$, we have $\angle COB=90^{\circ}$.
(i) Rotate $\triangle BOA$ 90 degrees clockwise around point $O$ to get $\triangle B'OA_1$.

At this point, $B'(-2,2)$ is the midpoint of $CO$, and point $A_1(4,-1)$.

Extend $OA_1$ to point $E_1$ such that $OE_1=2OA_1$, at this point, $E_1(8,-2)$ is the point that meets the condition.
(ii) Reflect $\triangle BOA$ across the $x$-axis to get $\triangle B'OA_2$, resulting in point $A_2(1,-4)$.

Extend $OA_2$ to point $E_2$ such that $OE_2=2OA_2$, at this point, $E_2(2,-8)$ is the point that meets the condition.
In summary, point $E(8,-2)$ or $(2,-8)$.