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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).

8) 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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Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
m = \frac{y_2-y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} =2
As it is given that the tangent is parallel to the chord, so their slopes are  equal
i.e.  slope of the tangent = slope of the chord
Given the equation of the curve is y = ( x-2)^2
\therefore \frac{dy}{dx} = 2(x-2) = 2
(x-2) = 1\\ x = 1+2\\ x=3
Now, when  x=3    y=(3- 2)^2 = (1)^2 = 1
Hence, the coordinates are \left ( 3,1) 
                 

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