91Ó°ÊÓ

The tangent line represents a straight line that touches a curve at a single point without crossing it. Considering the curve represented by the graph of a function, the tangent line at any given point provides the best linear approximation of the function near that point.
To illustrate this, let's take the curve given by the equation \[ y = \frac{1}{x} \] At any point \( P(a, b) \), the slope of the tangent line is obtained by evaluating the derivative of the function at \( P \). The slope determines how steep the tangent is and in what direction it moves.
Using the point-slope form of the equation, the tangent at \( P(a, b) \) can be written as: \[ y - b = m (x - a) \] where \( m \) is the slope of the tangent line. This form is useful in calculating the tangent line's intersection with axes.

Derivatives are a foundational concept in calculus, representing the rate at which a function changes at any point. It is the mathematical way of finding the slope of the curve at a point.
For the function \( y = \frac{1}{x} \), the derivative is: \[ y' = -\frac{1}{x^2} \] This derivative tells us that at any point \( x \), the slope of the tangent line is \( -\frac{1}{x^2} \).
By substituting \( x = a \) in this derivative, we find the slope at point \( P(a, \frac{1}{a}) \) which equals \( -\frac{1}{a^2} \). This negative slope indicates that the function decreases as \( x \) increases.

An isosceles triangle has at least two sides that are equal in length. In this exercise involving point \( A \) on the \( x \)-axis and point \( O \) at the origin, verifying the isosceles nature of triangle \( AOP \) involves comparing the lengths of sides.
Given:

• Point \( O(0,0) \) - the origin,
• Point \( A(a+1, 0) \) - on the x-axis,
• Point \( P(a, \frac{1}{a}) \) - on the curve.

The distance \( OA \) equals \( a+1 \). Using the distance formula, \( OP \) is also \( a+1 \).
Since \( OA = OP \), triangle \( AOP \) is confirmed to be isosceles, showcasing the balanced nature of this geometric shape around the point \( P \).

The area of a triangle gives us a measure of the space enclosed within its three sides. For triangle \( AOP \), where one side lies along the \( x \)-axis, calculating the area depends on the base and height.
The base \( OA \) is clearly \( a+1 \), with the height corresponding to the vertical distance from \( P \) to the \( x \)-axis, which is \( \frac{1}{a} \).
Using the formula for area: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] substitute the values to get: \[ \text{Area} = \frac{1}{2} \times (a+1) \times \left(\frac{1}{a}\right) = \frac{a+1}{2a} \] This result provides a concise measure of the space occupied by the triangle relative to point \( P \) and the tangent line's point of intersection with the \( x \)-axis.