Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes (if it is a hyperbola). $$ \frac{-x^{2}}{9}+\frac{y^{2}}{4}=1 $$

Step-by-Step Solution

Answer

It's a vertical hyperbola with vertices at (0, ±2), foci at (0, ±√13), and asymptotes y = ±(2/3)x.

Step 1: Identify the Type of Conic Section

The equation given is \(\frac{-x^{2}}{9} + \frac{y^{2}}{4} = 1\). It is structured as \(-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), which represents a hyperbola.

Step 2: Determine the Orientation of the Hyperbola

Since the positive term (\(y^2\)) is associated with the \(y\)-axis, the hyperbola is oriented vertically (opens upwards and downwards).

Step 3: Find the Vertices

For the hyperbola \(\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1\), the vertices are located at \( (0, \pm b) \). Here, \(b^2 = 4\), thus \(b = 2\). Therefore, the vertices are at \((0, 2)\) and \((0, -2)\).

Step 4: Find the Asymptotes

The equations of the asymptotes of a vertical hyperbola are given by \(y = \pm \frac{b}{a}x\). Here, \(a^2 = 9\) so \(a = 3\), and \(b = 2\). The slope \(\frac{b}{a} = \frac{2}{3}\). Thus, the asymptotes are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Step 5: Find the Foci

The foci of a vertical hyperbola are located at \((0, \pm c)\), where \(c^2 = a^2 + b^2\). Substituting the values, \(c^2 = 9 + 4 = 13\), and thus, \(c = \sqrt{13}\). The foci are therefore at \((0, \sqrt{13})\) and \((0, -\sqrt{13})\).

Step 6: Sketch the Graph

To sketch the graph: locate the center at the origin \((0, 0)\), plot the vertices at \((0, 2)\) and \((0, -2)\), draw the asymptotes through the origin with slopes \(\frac{2}{3}\) and \(-\frac{2}{3}\). Finally, sketch the hyperbola opening upwards and downwards between the asymptotes, passing through the vertices.

Key Concepts

Vertices of Hyperbola

In a hyperbola, the vertices are pivotal for understanding its shape and orientation. The vertices represent the closest points of the hyperbola to its center. In the case of the equation \(-\frac{x^2}{9} + \frac{y^2}{4} = 1\), we follow the general form \(-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, since the positive term lies with \(y^2\), the hyperbola is vertical.

• Orientation: A vertical hyperbola opens up and down, with its vertices along the y-axis.
• Finding the Vertices: The formula for the vertices is \((0, \pm b)\). From the equation, we have \(b^2 = 4\), which means \(b = 2\). Thus, the vertices are located at coordinates \((0, 2)\) and \((0, -2)\).

These vertices are essential in drawing the basic structure, as they dictate where the hyperbola starts curving away from its center.

Foci of Hyperbola

The foci are crucial in the hyperbola's definition and are used to describe its elongation. For a hyperbola, every point on its branches has a consistent distance difference to these two fixed points called foci.

• Finding the Foci: The foci are located off the center, extending along the axis of the hyperbola. For a vertical hyperbola, the foci formula is \((0, \pm c)\), where \(c\) is derived from \(c^2 = a^2 + b^2\).
• Calculating \(c\): Substitute the given values \(a^2 = 9\) and \(b^2 = 4\) into the formula, yielding \(c^2 = 13\). Therefore, \(c = \sqrt{13}\).
• Foci Coordinates: The points are \((0, \sqrt{13})\) and \((0, -\sqrt{13})\).

These foci help in constructing the exact curvature of the hyperbola, being integral to understanding the hyperbola's stretch and overall form.

Asymptotes of Hyperbola

Asymptotes provide a foundation for sketching hyperbolas accurately, showing how the branches of the hyperbola extend infinitely without bound. These diagonal lines frame the hyperbola but are never touched by its branches.

• Role of Asymptotes: They represent the directions in which the hyperbola opens, extending indefinitely along these guidelines.
• Equation of Asymptotes: For a vertical hyperbola, the asymptotes are determined by the formula \(y = \pm \frac{b}{a}x\).
• Calculating the Slope: In the equation \(-\frac{x^2}{9} + \frac{y^2}{4} = 1\), we find \(a = 3\) and \(b = 2\). Therefore, \(\frac{b}{a} = \frac{2}{3}\). The asymptote equations are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Understanding these asymptotes aids students in outlining the hyperbola's pathway and ensuring their sketches are accurately representative of the equation provided. They guide the form of the curve, aiding in precision during construction.