An answer of that angle which is constant and perpendicular to 2sqrt5 so get an answer of angle between tangential velocity directed along the shoreline direction and perpendicular to the sine of that angle between tangential velocity direction which is constant and shoreline is the radial beam [42pi radians sqrt5 km] min at the radial beam is component directed toward point for fixed radius.

yay! only just now did i understand the question.

let x be the distance from P along the shoreline.
and r the length of beam to shoreline (from lighthouse)

r^2 = 2^2 + x^2 = x^2 + 4

2r * (dr/dt) = 2x * (dx/dt)
r^2 = 1 + 4 = 5
r = sqrt(5)

2/r = cos y, 2 = rcos y
0 = (dr/dr) * cos y - sin y * (dy/dt) * r

dr/dt
= r(dy/dt)tan y
= sqrt(5) * (8pi) * (1/2)
= 4sqrt(5) pi

2r * (dr/dt) = 2x * (dx/dt)
dx/dt
= sqrt(5) * 4sqrt(5) pi / 1
= 20 pi km / minute