Question

Find the 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 the 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

Slope of chord joining (2,0) and (4,4) is $\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-2}=\frac{4}{2}=2$ ...(i)

Equation of given curve is $y=(x-2{)}^2$

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

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

Now, from Eq. (i), we get 2(x-2)=2 $\Rightarrow x-2=1\Rightarrow x=3$

When x=3, then $y=(x-2{)}^2$=1 [$\because$ Equation of curve is $y=(x-2{)}^2$]

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