Find a point on the curve y=(x-2)^{2} at whic | vedclass

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(A) If a tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$,then the slope of the tangent must be equal to the slope of the chord

Vedclass · Accepted Answer

$(3,1)$

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(A) If a tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$,then the slope of the tangent must be equal to the slope of the chord.The slope of the chord joining $(x_1, y_1) = (2,0)$ and $(x_2, y_2) = (4,4)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} = 2$.Now,the slope of the tangent to the curve $y = (x-2)^2$ at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 2(x-2)$.Since the tangent is parallel to the chord,we equate the slopes: $2(x-2) = 2$.Dividing by $2$,we get $x-2 = 1$,which implies $x = 3$.To find the corresponding $y$-coordinate,substitute $x=3$ into the curve equation: $y = (3-2)^2 = 1^2 = 1$.Thus,the required point on the curve is $(3,1)$.

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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