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 $(4,4)$.

Answer 1

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, $\frac{dy}{dx}=2(x-2)$

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 $(3,1).$