+845 The point A lies on the curve with the equation = 𝑥^1/2 . The tangent to this curve at A is parallel to the line 3𝑦 −2𝑥 = 1 . Find the equation of this tangent at A.
+128448 The line will have the slope = 2/3
The derivative of x^(1/2) = 1 / ( 2√x)
So....to find the x coordiniate of A
2/3 = 1 / ( 2√x)
( 2√x) = 3/2 square both sides
4x = 9/4
x = 9/16
So....the y coordinate of A is (9/16)^1/2 = 3/4
And the equation of the tangent line is
y = (2/3) (x - 9/16) + 3/4
y = (2/3)x - 9/24 + 3/4
y = (2/3)x + 9/24
24y = 16x + 9
16x - 24y = -9
