At what point on the curve y=sqrt(1+2x) is the tangent line perpendicular to the line 6x+2y=1

The slope of the given line is -3. The slope of the tangent line must be 1/3. 

y = √(1 + 2x) 

Tangent line: 
y = 1/3x + c, where c is still unknown 

Substitute for y in the curve equation. 
1/3x + c = √(1 + 2x) 
(1/3x + c)² = 1 + 2x 
1/9x² + 2/3cx + c² = 1 + 2x 
x² + 6cx + 9c² = 9 + 18x 
x² + (6c - 18)x + (9c² - 9) = 0 

discriminant = 0 
(6c - 18)² - 4(1)(9c² - 9) = 0 
(c - 3)² - (c² - 1) = 0 
-6c + 10 = 0 
c = 5/3 

x² + [6(5/3) - 18]x + [9(5/3)² - 9] = 0 
x² - 8x + 16 = 0 
(x - 4)² = 0 
x = 4 

y = √[1 + 2(4)] = 3 

Point of tangency: (4, 3)

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