91Ó°ÊÓ

A hyperbola is a type of conic section formed by intersecting a double cone. It features two symmetric curves that open away from each other. The standard form of a hyperbola can be represented as \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). This equation describes a hyperbola that opens along the y-axis.
In the given exercise, the equation \( \frac{y^2}{4} - \frac{x^2}{2} = 1 \) follows this pattern, indicating a hyperbola. The hyperbola has its center at the origin, and its asymptotes will guide the direction in which the curves open. Studying hyperbolas involves understanding how they differ from other conics like ellipses and parabolas in terms of structure and properties.

When working with hyperbolas, it’s often necessary to use implicit differentiation to find the slope of a tangent line. Implicit differentiation is a technique used when you have an equation involving both \( x \) and \( y \) that cannot be easily solved for \( y \) in terms of \( x \).
In this exercise, implicit differentiation is applied to \( \frac{y^2}{4} - \frac{x^2}{2} = 1 \). By differentiating each term with respect to \( x \), including the application of the chain rule for \( y^2 \), we find the derivative to be \( 0 = \frac{y \cdot y'}{2} - x \). Solving for \( y' \), we get \( y' = \frac{2x}{y} \), which represents the slope of the tangent line to the hyperbola at any point \( (x, y) \).

Normal lines are lines that are perpendicular to tangent lines at the point of tangency. Therefore, we can calculate the slope of a normal line as the negative reciprocal of the tangent line's slope.
For instance, if the slope of a tangent line is \( \frac{4}{3} \), the slope of the respective normal line will be \( -\frac{3}{4} \). This relationship is crucial in geometry and applications involving any perpendicularity principle.
In the context of the exercise, after finding the slopes of the tangents, the slopes of the normals are determined subsequently: \(-\frac{3}{4}\) and \(\frac{3}{4}\). Understanding how these slopes relate helps in constructing the equations for the normal lines.

The slope of the tangent line to a curve at a given point is determined by the derivative of the curve's equation evaluated at that point. In this hyperbola problem, we have already found the derivative using implicit differentiation as \( y' = \frac{2x}{y} \).
To find the slope at specific points, substitute the given \( x \) and the corresponding values of \( y \) into this derivative equation. When \( x = 4 \) and \( y = 6 \), the slope \( y' = \frac{4}{3} \). Similarly, for \( y = -6 \), the slope \( y' = -\frac{4}{3} \).
These calculated slopes are vital in drawing the tangent lines and determining their exact expressions in linear form.

To write the equation of a line when you know its slope and a point it passes through, you can use the point-slope form given by \( y - y_1 = m(x - x_1) \). This form is particularly useful for finding equations of tangent and normal lines once you've determined the slope and point of tangency.
For example, with a slope \( m = \frac{4}{3} \) and a point \( (4, 6) \), the equation of the tangent line becomes \( y - 6 = \frac{4}{3}(x - 4) \). By solving and simplifying this equation, we reach the standard line equation form, which describes the tangent line directly. Using this template makes the process of finding such equations simple and intuitive.