Find a 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).

$(3,1)$

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

Slope of chord joining points (2, 0) and (4, 4) = $\frac{4-0}{4-2}= 2$.

Let $P(x_1,y_1)$ be a point on the given curve at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

The given curve is $y = (x-2)^2$   ...(i)

Differentiating (i) w.r.t. x, we get

$\frac{dy}{dx} = 2(x-2)$.

∴ Slope of tangent at $P(x_1, y_1) =\left(\frac{dy}{dx}\right)_P=2(x_1-2)$.

Since the tangent to the given curve at $P(x_1,y_1)$ is parallel to the chord,

slope of tangent at P = slope of chord

$⇒2(x_1-2)=2⇒ x_1-2=1⇒x_1=3$.

As $P(x_1,y_1)$ lies on the curve (i), $y_1 = (x_1-2)^2$

$⇒ y_1 = (3-2)^2 ⇒y_1 = 1$.

Hence, the required point is (3, 1).