# 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).

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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