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 (x, y) 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).