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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). |
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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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