Problem 33 - Precalculus

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Use a graphing utility to graph the hyperbola and its asymptotes. $$2 x^{2}-3 y^{2}=6$$

Step-by-Step Solution

Answer

The center of the hyperbola is at (0, 0), the vertices are at \(\pm \sqrt{3}\) or approximately \(\pm 1.73\), the foci are at \(\sqrt{5}\) or approximately \(\pm 2.24\), and the equations of the asymptotes are \(y = \pm (\frac{\sqrt{2}}{\sqrt{3}})x\).

Step 1: Rewrite the Equation

First, rearrange the given equation \(2 x^{2}-3 y^{2}=6\) into standard form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). Divide all terms by 6 to begin the process, obtaining: \(\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1\).

Step 2: Identify the Center

The hyperbola standard form also specifies the center coordinates \(h, k\). In this case, since the format is \(x^2 - y^2 = 1\), the center is at the origin, which is the point (0, 0).

Step 3: Identify the Vertices

The vertices are at \(\pm a\), so in the case of the given hyperbola, the vertices are at \(\pm \sqrt{3}\) or approximately \(\pm 1.73\) on the x-axis.

Step 4: Identify the Foci

The foci are found using the formula \(c= \sqrt{a^{2} + b^{2}}\), where \(a = \sqrt{3}\) and \(b = \sqrt{2}\). Calculate to find \(c = \sqrt{3 + 2}\) which simplifies to \(c = \sqrt{5} \approx \pm 2.24\) on the x-axis.

Step 5: Identify the Asymptotes

The equations for the asymptotes are given by \(\pm \frac{y}{b} = \frac{x}{a}\). Given \(a = \sqrt{3}\) and \(b = \sqrt{2}\), the equations for the asymptotes are \(y = \pm (\frac{\sqrt{2}}{\sqrt{3}})x\).

Key Concepts

Standard Form of Hyperbola

The standard form of a hyperbola is crucial for identifying its center, vertices, foci, and asymptotes. A hyperbola can be expressed as either \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) or \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \). These correspond to horizontal and vertical hyperbolas, respectively.

In this exercise, the given equation is \( 2x^2 - 3y^2 = 6 \). To express it in standard form, divide every term by 6, resulting in \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \). Notice, the term with \( x \) is positive, indicating a horizontal hyperbola.

The center of the hyperbola can be directly revealed from this form. Since it’s \( x^2 \) and \( y^2 \) with no linear terms for \( x \) or \( y \), the center is at the origin: \((h, k) = (0,0)\). From here, you can systematically find the vertices, foci, and asymptotes.

Vertices of Hyperbola

Vertices provide crucial information on the extent of a hyperbola along its central axis. With the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are found \( \pm a \) units away from the center, which is along the x-axis for this equation.

In the equation \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \), we identify \( a^2 = 3 \). Therefore, \( a = \sqrt{3} \) which is approximately 1.73. Thus, the vertices are located at \( (\pm \sqrt{3}, 0) \) or approximately at \( (\pm 1.73, 0) \).

These points mark the widest points of the hyperbola along the x-axis, giving it its characteristic shape of narrowly opening curves.

Foci of Hyperbola

The foci are not just randomly placed; they anchor the hyperbola's curves. You can find them with the formula \( c = \sqrt{a^2 + b^2} \). This formula is derived from the hyperbola's geometry, highlighting the constant difference in distances from any point on the hyperbola to these foci.

For \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \), \( a = \sqrt{3} \) and \( b = \sqrt{2} \). Plug these values into the formula: \( c = \sqrt{3 + 2} = \sqrt{5} \). Thus, \( c \approx 2.24 \).

Therefore, the foci are at \( (\pm \sqrt{5}, 0) \) or approximately \( (\pm 2.24, 0) \) on the x-axis, lying farther away from the hyperbola’s center than the vertices.

Asymptotes of Hyperbola

Asymptotes are invisible guides that tell how a hyperbola behaves as it extends away from the center. These lines are defined by the slopes \( \pm \frac{b}{a} \), where \( a \) and \( b \) correspond to the hyperbola's standard form.

For the given equation \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \), we have \( a = \sqrt{3} \) and \( b = \sqrt{2} \). Therefore, the asymptotes are given by \( y = \pm \frac{\sqrt{2}}{\sqrt{3}} x \).

These lines do not intersect the hyperbola but closely approach its curves as \( x \) approaches infinity, giving a framework for the hyperbola's path, depicting its direction of opening and closure.