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

Solution: Equation of the curve is,
$y={\left(x-2\right)}^2$
$\therefore \frac{dy}{dx}=2\left(x-2\right)$
$\therefore$ Slope of the tangent$\left(m_1\right)=\frac{dy}{dx}=2\left(x-2\right)$
Now, slope of the chord joining points $\left(2,0\right)$ and $\left(4,4\right)=\frac{4-0}{4-2}=2$
As, chord is parallel to the tangent, so their slopes will be equal.
$\therefore 2\left(x-2\right)=2$
$\Rightarrow x=3$
$\therefore y={\left(3-2\right)}^2=1$
So, the required point is $\left(3,1\right)$.