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Question

Find a point on the curve y=(x2)2 at which the tangent is parallel to the chord joining the points (2,0) and (4,4).

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Solution

y=(x2)2
dydx=d((x2)2)dx=2(x2)
slope of tangent =2x4
Slope of line joining (2,0) and (4,4) =4042=2
The tangent is parallel to this line
their slopes are equal
2x4=2 2x=6
x=3
and y=(32)2=1
Thus the point is (3,1)

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