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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