The point on the curve y=square root of 2x+1 at which the tangent is parallel to the line x-3y=6 is ?

Answer 1

Slope of tangent line m is equel the slope of given line. To get slope of given line let solve given equation for y:

x - 6 = 3y, y = 1/3x - 2. So slope line is 1/3 and m = 1/3.

From other side slope equel derivative at tangent point.

y = (2x + 1)^1/2, then y' = 1/2·(2x + 1)^-1/2·2 = (2x + 1)^-1/2.

But y' = m = 1/3. We get equation for x-coordinate of tangent point: (2x + 1)^-1/2 = 1/3. Rasing both side to power -2. We get 2x + 1 = 9, 2x = 8, x = 4.

Substitude this 4 in equation of line to get y-coordinate of tangent point

y = (2·4 + 1)^1/2 = 9^1/2 = 3.

Answer: tangent point is (4, 3)