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

$y=(x-2{)}^2$

$\frac{dy}{dx}=\frac{d\left(\right(x-2{)}^2)}{dx}=2(x-2)$

$\therefore$ slope of tangent =$2x-4$

Slope of line joining $(2,0)$ and $(4,4)$ $=\frac{4-0}{4-2}=2$

The tangent is parallel to this line

$\therefore$ their slopes are equal

$2x-4=2$ $\Rightarrow 2x=6$

$\therefore x=3$

and $y=(3-2{)}^2=1$

Thus the point is $(3,1)$