Answer:
the diameter of the outside edge of the receiver is
![8√(18) \ \ \ inches](https://img.qammunity.org/2021/formulas/physics/college/lk64omdfmwtjkihys2fsbypr2mz9s0ekpx.png)
Step-by-step explanation:
From the schematic free body diagram illustrating what the question is all about below;
Let represent A to be the vertex where the receiver is being placed
S to be the focus
BP to be equal to r (i.e radius of the outer edge)
BC to be 2 r (i.e the diameter)
Given that AS = 4 in and AP is 18 in
Let AP be x- axis and AY be y -axis
A=(0,0)
S=(4,0) = (0,0)
So that the equation of the parabolic path of the receiver will be:
![y^2 =4 ax \\ \\ y^2 = 4*4*x \\ \\ y^2 = 16x](https://img.qammunity.org/2021/formulas/physics/college/r4gpnmn13gsfy4evvzfexexqd1ceite0cf.png)
B = (AP, BP)
B = (18, r)
B lies y² = 16 x
r² = 16 x
r² 16 × 18
![r = √(16*18 ) \\ \\ r = 4√(18)](https://img.qammunity.org/2021/formulas/physics/college/c8gwk43v4o3837iq8y9sj9ujqux9lf524z.png)
Diameter BC = 2r
![2* 4√(18) \\ \\= 8√(18 ) \ \ \ inches](https://img.qammunity.org/2021/formulas/physics/college/u3fvkn4fg1eevrf1ahfb2aeyn154a23biw.png)