Find a point on the curve $y = {\left( {x - 2} \right)^2}$ at which the tangent is parallel to the chord joining the points $\left( {2,0} \right)$ and $\left( {4,4} \right)$.

Equation of given curve is $y = {\left( {x - 2} \right)^2}$ …(i)

$\Rightarrow \cfrac{{dy}}{{dx}} = 2(x - 2)$

Slope of chord joining the points $\left( {2,0} \right)$ and $\left( {4,4} \right)$ is
$\cfrac{{4 - 0}}{{4 - 2}} = \cfrac{4}{2} = 2$

For the points at which tangent is parallel to the chord joining points $\left( {2,0} \right)$ and $\left( {4,4} \right)$, we must have

$\cfrac{{dy}}{{dx}} =$ slope of the chord

$\Rightarrow 2\left( {x - 2} \right) = 2 \Rightarrow x - 2 = 1 \Rightarrow x = 3$

When $x = 3$, then from (i), we get $y = {\left( {3 - 2} \right)^2} = 1$

Therefore the required point is $\left( {3,1} \right)$