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Question

Find a point on the curve y=x3+1, where the tangent is parallel to the chord joining (1,2) and (3,28).

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Solution

It is given that tangent is parallel to chord

slope of tangent = slope of chord

Let m1= slope of tangent and m2=y2y1x2x1= slope of line joining 2 points

y=x3+1 m2=262=13

dydx=3x2=m1

m1=m2

3x2=13

x=133

y=(133)3+1

y=133133+1

Points on the curve are (133,1313+3333)

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