Problem 14 - Precalculus Mathematics for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$9 x^{2}-4 y^{2}=36$$

Step-by-Step Solution

Answer

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

Step 1: Write the Equation in Standard Form

The equation of the hyperbola is given as \(9x^2 - 4y^2 = 36\). To convert it to the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), divide the entire equation by 36. This yields \(\frac{x^2}{4} - \frac{y^2}{9} = 1\).

Step 2: Identify Parameters a and b

From the standard form \(\frac{x^2}{4} - \frac{y^2}{9} = 1\), recognize that \(a^2 = 4\) and \(b^2 = 9\). Thus, \(a = 2\) and \(b = 3\).

Step 3: Find the Vertices

For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are at \((\pm a, 0)\). So, the vertices are \((\pm 2, 0)\) or \((-2, 0)\) and \((2, 0)\).

Step 4: Calculate the Foci

The distance of the foci from the center along the x-axis is given by \(c\), where \(c^2 = a^2 + b^2\). Calculate \(c\) by \(c = \sqrt{4 + 9} = \sqrt{13}\). The foci are at \((\pm \sqrt{13}, 0)\).

Step 5: Determine the Asymptotes

The asymptotes for a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by \(y = \pm \frac{b}{a}x\). Thus, the asymptotes are \(y = \pm \frac{3}{2}x\).

Step 6: Sketch the Graph

Plot the vertices \((-2, 0)\) and \((2, 0)\). Then, plot the asymptotes \(y = \frac{3}{2}x\) and \(y = -\frac{3}{2}x\). Draw the branches of the hyperbola opening left and right (as the x-term is positive in the standard form).

Step 7: Conclude the Features

The vertices are \((-2, 0)\) and \((2, 0)\), the foci are located at \((-\sqrt{13}, 0)\) and \((\sqrt{13}, 0)\), and the asymptotes are \(y = \frac{3}{2}x\) and \(y = -\frac{3}{2}x\).

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. These points are crucial because they represent the closest and farthest points horizontally from the center of the hyperbola in one direction. For a hyperbola in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are located at \((\pm a, 0)\). In our example, \(\frac{x^2}{4} - \frac{y^2}{9} = 1,\) we identify \(a^2 = 4\), which gives \(a = 2\).Therefore, the vertices are \((-2, 0)\) and \((2, 0)\). These points mark where the hyperbola's branches are closest and farthest apart along the x-axis. Think of drawing a line segment through \((-2, 0)\) and \((2, 0)\) which acts as the line on which the hyperbola’s two branches stem outward.

Foci of Hyperbola

The foci of a hyperbola are two critical points located inside each branch of the hyperbola. They are used to define the hyperbola's properties, as the difference in the distances from any point on the hyperbola to the foci remains constant. For a hyperbola in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\) the distance \(c\) from the center to each focus is determined by the formula \(c^2 = a^2 + b^2\). In our case, \(a = 2\) and \(b = 3\), so \(c = \sqrt{4 + 9} = \sqrt{13}\). This gives two foci at \((-\sqrt{13}, 0)\) and \((\sqrt{13}, 0)\). These points lie along the x-axis, extending outside the vertices. The significance of these points is pivotal in graphing, as they help in understanding the openness and orientation of the hyperbola's branches.

Asymptotes of Hyperbola

The asymptotes of a hyperbola are straight lines that the branches of the hyperbola approach but never actually meet. They provide a boundary that determines the direction in which the branches of the hyperbola extend. For a standard hyperbola given by\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\) the asymptotes are of the form \(y = \pm \frac{b}{a}x\). For our hyperbola, \(a=2\) and \(b=3\), so the asymptotes are \(y = \frac{3}{2}x\) and \(y = -\frac{3}{2}x\). These asymptotes create a kind of crossing diagonal envelope within which the branches of the hyperbola sit, providing a visual guide for the hyperbola's shape and orientation. Understanding the role of asymptotes is crucial for sketching the hyperbola accurately. They ensure that the curves of the hyperbola open symmetrically and provide a geometrical boundary for the graph.