91Ó°ÊÓ

The equation of a tangent line is essential in calculus, as it represents the line that just "touches" a curve at a particular point. For our given curve, described by the equation \( y^2 = kx \), we want to determine the equation of the tangent line at a specific point \((x_0, y_0)\). The tangent line shares the same slope as the curve at the point of tangency. This is found by calculating the derivative of the curve, which is \( \frac{dy}{dx} \), at that particular point.
The tangent line equation gives us a linear approximation of the curve near the point \((x_0, y_0)\). This allows for simpler calculations and a better understanding of the behavior of the curve near that point. In this case, the equation we derive will be simplified to show that it matches the form \( y_0 y = \frac{1}{2}k(x + x_0) \), which reflects the curve's characteristics accurately at \((x_0, y_0)\).

Calculating the derivative is a critical step in finding the tangent line's equation. We begin with implicit differentiation since our original equation, \( y^2 = kx \), involves both \( x \) and \( y \). Implicit differentiation allows us to differentiate both sides of the equation with respect to \( x \) without explicitly solving for \( y \).

Upon applying implicit differentiation, we differentiate \( y^2 \) to get \( 2y \frac{dy}{dx} \) and \( kx \) becomes \( k \). By equating these, we have \( 2y \frac{dy}{dx} = k \). Solving this equation for \( \frac{dy}{dx} \), we find that \( \frac{dy}{dx} = \frac{k}{2y} \), which is the derivative or the slope of the tangent line to the curve.

• This result shows how changes in \( x \) impact changes in \( y \) along the curve.
• This calculation is fundamental, as the slope \( m \) is later used in the tangent line formula.

The point-slope form of a line is a powerful tool for writing the equation of a line when you know its slope and a point on the line. It is expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is a known point.
In our problem, after calculating the derivative \( \frac{dy}{dx} = \frac{k}{2y_0} \), this derivative acts as the slope \( m \). Using the point-slope form, we plug in the values \( y_0 \) and \( x_0 \) from the point \((x_0, y_0)\) and the slope \( m \).

This results in the equation \( y - y_0 = \frac{k}{2y_0}(x - x_0) \). Through rearranging and simplifying, following the steps, this eventually substantiates the tangent line equation \( y_0 y = \frac{1}{2}k(x + x_0) \)

• The point-slope form is helpful because it makes it straightforward to determine the equation of a tangent, given a derivative and a specific point.
• It directly illustrates the relationship between the slope and point on the curve.