Find a point on the curve

Question:

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 $(4,4)$.

Solution:

If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.

The slope of the chord is $\frac{4-0}{4-2}=\frac{4}{2}=2$.

Now, the slope of the tangent to the given curve at a point (xy) is given by,

$\frac{d y}{d x}=2(x-2)$

Since the slope of the tangent = slope of the chord, we have:

$2(x-2)=2$

$\Rightarrow x-2=1 \Rightarrow x=3$

When $x=3, y=(3-2)^{2}=1$

Hence, the required point is (3, 1).

 

 

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