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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). |
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Answer» `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). |
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