derivatives

Find the point on curve y = (2x - 1)^ 0.5 when the tangent line is parallel to line x - 3y = 16.


Let (x1, y1) be the point required.

y = (2x - 1)^0.5
dy/dx = [0.5(2x - 1)^(-0.5)]*2
dy/dx = (2x - 1)^(-0.5)

Slope of the tangent:
(2x1 - 1)^(-0.5) = -1/(-3)
(2x1 - 1)^(-0.5) = 1/3
(2x1 - 1)^0.5 = 3
2x1 - 1 = 9
x1 = 5

(x, y) lies on the curve:
y1 = [2(5) - 1]^0.5
y1 = 3

Hence, the point required is (5, 3).