Find the focus, directrix, focal diameter, vertex and axis of symmetry for the parabola
$$43.6x=y^2$$

Focus
Directrix
Focal diameter =
Vertex
Axis of symmetry

Step 1 ：The given equation is in the form of $y^2=4ax$, where $4a=43.6$.

Step 2 ：Calculate the value of $a$, $a=\frac{43.6}{4}=10.9$.

Step 3 ：The focus of the parabola is at $(a,0)$, so the focus is at $(10.9,0)$.

Step 4 ：The directrix of the parabola is $x=-a$, so the directrix is $x=-10.9$.

Step 5 ：The focal diameter of the parabola is $4a$, so the focal diameter is $43.6$.

Step 6 ：The vertex of the parabola is at the origin $(0,0)$.

Step 7 ：The axis of symmetry of the parabola is the y-axis $x=0$.

Step 8 ：$\boxed{{\displaystyle \text{Final Answer: The focus of the parabola is at }(10.9,0),\text{ the directrix is }x=-10.9,\text{ the focal diameter is }43.6,\text{ the vertex is at the origin }(0,0),\text{ and the axis of symmetry is the y-axis }x=0}}$.