Problem 1 - Algebra and Trigonometry with Analytic Geometry

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$

Step-by-Step Solution

Answer

Vertices: (3, 0) and (-3, 0); Foci: (\(\sqrt{13}\), 0) and (-(\(\sqrt{13}\), 0); Asymptotes: \(y = \pm \frac{2}{3}x\).

Step 1: Identify the Hyperbola Form

The given equation is \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \). This is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for a hyperbola centered at the origin with a horizontal transverse axis.

Step 2: Find Vertices

For a hyperbola with equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at \((\pm a, 0)\). Here, \(a^2 = 9\), so \(a = 3\). Thus, the vertices are \((3, 0)\) and \((-3, 0)\).

Step 3: Calculate the Foci

The foci are located \((\pm c, 0)\), where \(c^2 = a^2 + b^2\). Here, \(c^2 = 9 + 4 = 13\), giving \(c = \sqrt{13}\). Therefore, the foci are \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\).

Step 4: Determine the Asymptotes

The equations of the asymptotes for this hyperbola are given by \(y = \pm \frac{b}{a}x\). Here, \(b^2 = 4\), so \(b = 2\). Therefore, the asymptotes are \(y = \pm \frac{2}{3}x\).

Step 5: Sketch the Graph

Draw the hyperbola by plotting vertices and drawing the asymptotes as dashed lines through the origin with slopes \(\frac{2}{3}\) and \(-\frac{2}{3}\). The hyperbola opens horizontally with vertices at \((3,0)\) and \((-3,0)\), and the foci are near these vertices at \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\).

Key Concepts

Vertices of a Hyperbola

Understanding the vertices of a hyperbola is crucial as they mark the turning points of the open curves. The vertices are located on the transverse axis of the hyperbola, which is the axis that runs through the center and along the line that cuts through the open sides of the hyperbola.
For a hyperbola with the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the center is at the origin \((0,0)\), and the transverse axis is horizontal.
Here, the parameter \(a^2\) from the equation is 9, making \(a = 3\). Therefore, the vertices are positioned at \((3, 0)\) and \((-3, 0)\).
These points help form the basic shape of the hyperbola and are essential when sketching its graph.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points that the curve surrounds. Unlike the vertices, the foci lie further out along the transverse axis. The importance of the foci is in the hyperbolic shape’s definition, as the difference in distances from any point on the hyperbola to these foci is constant.
To find the foci for a standard hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), calculate \(c\) where \(c^2 = a^2 + b^2\).
With \(a^2 = 9\) and \(b^2 = 4\), we find \(c^2 = 13\), giving \(c = \sqrt{13}\).
Thus, the foci are \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\). These points, typically outside the vertices on the graph, help in achieving accurate representation.

Asymptotes of a Hyperbola

Asymptotes are straight lines that the curve approaches but never touches. For a hyperbola, they are crucial for sketching the graph accurately, as the curve moves infinitely close to these lines as you move further from the center.
The equations of the asymptotes for a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \(y = \pm \frac{b}{a}x\).
Here, \(b^2 = 4\), hence \(b = 2\), and with \(a = 3\), the asymptotic lines are \(y = \pm \frac{2}{3}x\).
These asymptotes aid in guiding the overall shape of the hyperbola and are often drawn as dashed lines in graphical depictions.

Equation of a Hyperbola

The equation of a hyperbola is formulated differently depending on the orientation of its transverse axis. A horizontal transverse axis, like in the given form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), dictates that the hyperbola opens sideways.
This standard form allows us to extract key parameters quickly: \(a\) and \(b\), which denote the distances from the center to the vertices and co-vertices, respectively.
By analyzing this equation, we determine \(a = 3\) and \(b = 2\). From here, further details like vertices, foci, and asymptotes can be calculated, each critical for plotting the hyperbola accurately on a graph.

Graphing Hyperbolas

Graphing a hyperbola involves plotting its elements and understanding its general shape. Begin by marking the center, located at the origin in standard cases. Plot the vertices to indicate the course of the curve; for our example, they are \((3, 0)\) and \((-3, 0)\).
Next, draw the asymptotes. Sketching these as dashed lines with slopes \(\pm \frac{2}{3}\) helps set the boundaries within which the hyperbola lies. Included are foci points \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\), usually marked for reference.
Finally, sketch the hyperbolic curves, ensuring they bow outwards from the center and approach the asymptotes within either quadrant without crossing them. This method provides a comprehensive graph of the hyperbola.