The point on the curve $y = (x-2)^2$ at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4) is:

(3, 1)

The correct answer is Option (4) → (3, 1)

Given curve: $y = (x-2)^2$

Chord joining points (2,0) and (4,4):

Slope of chord: $m = \frac{4 - 0}{4 - 2} = \frac{4}{2} = 2$

Derivative of $y$: $y' = \frac{d}{dx} (x-2)^2 = 2(x-2)$

Set slope of tangent = slope of chord:

$2(x-2) = 2 \Rightarrow x-2 = 1 \Rightarrow x = 3$

Corresponding $y$ value: $y = (3-2)^2 = 1$

Answer: (3, 1)