Question

Find a point on the curve y=(x-2)² at which the tangent is parallel to the line joining the points (2,0) and (4,4)

Answer 1

Answer:

Hey mate here is your answer is in image

Answer 2

Hey mate !!!

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 =

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)

Hope this helps you!!!