# Find a point on the curve y = (x – 3)2,where the tangent is parallel to the chord joining the points (3, 0) and (4, 1).

Given: Equation of curve, y = (x – 3)2

Firstly, we differentiate the above equation with respect to x, we get

[using chain rule]

Given tangent to the curve is parallel to the chord joining the points (3,0) and (4,1)

i.e.

2x – 6 = 1

2x = 7

Put in y = (x – 3)2 , we have

Hence, the tangent to the curve is parallel to chord joining the points (3,0) and (4,1) at

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