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).

Explanation

`y=(x-2)^(2)`…(1)
`rArr (dy)/(dx) = 2 (x-2)`
Slope of chord ` = (4-0)/(4-2) = 2`
According to the problem, for parallel lines,
` 2(x-2) = 2`
` rArr x= 3`
From equation (1)
` y=(3-2)^(2) = 1`
Therefore, required point is (3, 1).