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

given curve , y = (x - 2)²
A/C to question,
the tangent of the curve is parallel to the chord joining the points (2,0) and (4,4).
e.g., slope of tangent = slope of the chord joining the points (2,0) and (4,4)
slope of the chord joining the points (2,0) and (4,4) = (4 - 0)/(4 -2) = 4/2 = 2
so, slope of tangent = 2

we know slope of the tangent of the curve at (a,b) is 1st order derivatives of the curve at that given point . e.g., slope of tangent = $\bf{\frac{dy}{dx}|_{(a,b)}}$

so, differentiate y with respect to x ,
dy/dx = 2(x - 2)
Let required point is (a,b)
at (a,b) , dy/dx = 2(a - 2)
hence, slope of tangent = 2(a - 2)

now, slope of tangent = slope of chord
2(a - 2) = 1 => a = 3
put a is not other than x - coordinate of point
so, x = 3, put it in y = (x -2)²
e.g.,y = (3-2)² = 1

hence, required point is (3,1)