Problem:
A tangent is drawn to the curve y=8/x² at a point A(x₁,y₁) where x₁= 2. The tangent cuts the x-axis at point B. Find the scalar product of the vector AB and OB.

Approach:
To solve this problem, we need to find the coordinates of point B and then calculate the scalar product of vectors AB and OB.

Solution:

Step 1: Find the slope of the tangent:
The slope of the tangent to the curve y=8/x² at any point (x,y) is given by the derivative of the curve at that point.
Differentiating y=8/x² with respect to x, we get:
(dy/dx) = (-16/x³)

Substituting x₁=2, we get:
(dy/dx) = (-16/2³) = -2

So, the slope of the tangent at point A is -2.

Step 2: Find the equation of the tangent:
We know that the equation of a line with slope m passing through a point (x₁,y₁) is given by:
(y-y₁) = m(x-x₁)

Substituting the values of x₁=2, y₁=8/2²=2 in the above equation, we get:
(y-2) = -2(x-2)

Simplifying the equation, we get:
y = -2x + 6

So, the equation of the tangent is y = -2x + 6.

Step 3: Find the x-coordinate of point B:
To find the x-coordinate of point B, we need to solve the equation of the tangent for y=0.
Substituting y=0 in the equation y = -2x + 6, we get:
0 = -2x + 6
2x = 6
x = 3

So, the x-coordinate of point B is 3.

Step 4: Find the coordinates of point B:
We already know that the x-coordinate of point B is 3. To find the y-coordinate of point B, we substitute x=3 in the equation y=8/x²:
y = 8/3² = 8/9

So, the coordinates of point B are (3, 8/9).

Step 5: Find the vectors AB and OB:
The vector AB can be calculated using the coordinates of points A and B as:
AB = B - A = (3, 8/9) - (2, 2) = (1, 8/9 - 2) = (1, -10/9)

The vector OB is simply the position vector of point B, which is given by:
OB = B - O = (3, 8/9) - (0, 0) = (3, 8/9)

Step 6: Find the scalar product of vectors AB and OB:
The scalar product of two vectors is given by the dot product of the vectors.
The dot product of vectors AB and OB is given by:
AB · OB = (1)(3) + (-10/9)(8/9) = 3 - 80/81 = 243/