Find the maximum distance from the curve y=2√2x to the point (6, 0).

Answer 1

Check your question, you must mean 'minimum' distance.

Method 1:
slope of the given line is 2√2 or √8
so the slope of the line from (6,0) to the line must be -1/√8

the equation of that perpendicular is y = (-1/√8)x + b, with (6,0) on it, so
0 = (-1/√8)(6) + b
b = 6/√8

where do they intersect?
(-1/√8)x + 6/√8 = √8x
multiply each term by √8
-x + 6 = 8x
x = 2/3 , then y = 2√8/3
distance between (2/3 , 2√8/3) and (6,0)
= √( (16/3)^2 + (-2√8/3)^2)
= √(256/9 + 32/9)
= √288/3
= 12√2/3 = 4√2

method 2: using the formula for distance from a point to a given line.
line: √8x - y = 0 , point (6,0)
= (6√8 - 0 + 0)/√(√8^2 + (-1)^2)
= 6√8/3 = 2√8
= 4√2