Answer:
the diameter of the outside edge of the receiver is [tex]8\sqrt{18} \ \ \ inches[/tex]
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:
[tex]y^2 =4 ax \\ \\ y^2 = 4*4*x \\ \\ y^2 = 16x[/tex]
B = (AP, BP)
B = (18, r)
B lies y² = 16 x
r² = 16 x
r² 16 × 18
[tex]r = \sqrt{16*18 } \\ \\ r = 4\sqrt{18}[/tex]
Diameter BC = 2r
[tex]2* 4\sqrt{18} \\ \\= 8\sqrt{18 } \ \ \ inches[/tex]