if 3x4y0 is tangent in the first quadrant to the curve y x3

if 3x-4y=0 is tangent (in the first quadrant) to the curve y = x^3 + K
Find the value of K

Solution

y=x^3+k
=> dy/dx = 3x^2

If 3x-4y=0 then y=3x/4, so it must be tangent at the point (x,y) where

3x^2 = 3/4 => x=1/2 (taking positive root as in first quadrant)
=>y=3/8

but this point lies on the curve y=x^3+k
=> 3/8 = 1/8 + k
=> k= 2/8 = 1/4