1.

Find 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).

Answer» Equation of the curve is,
`y = (x-2)^2`
`:. dy/dx = 2(x-2)`
`:.` Slope of the tangent`(m_1) = dy/dx = 2(x-2)`
Now, slope of the chord joining points `(2,0)` and `(4,4) = (4-0)/(4-2) = 2`
As, chord is parallel to the tangent, so their slopes will be equal.
`:. 2(x-2) = 2`
`=>x = 3`
`:. y = (3-2)^2 = 1`
So, the required point is `(3,1)`.


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