Problem 24 - Precalculus: Mathematical for Calculus

Question

Use a graphing device to graph the hyperbola. $$3 y^{2}-4 x^{2}=24$$

Step-by-Step Solution

Answer

Graph the hyperbola by plotting vertices, co-vertices, and asymptotes, then sketch the branches.

Step 1: Rewrite the Equation

The given equation is \(3y^2 - 4x^2 = 24\). Rewrite the equation to match the standard form of a hyperbola, which is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Divide every term by 24 to obtain the equation \(\frac{y^2}{8} - \frac{x^2}{6} = 1\).

Step 2: Identify the Center

Since the equation is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the center of the hyperbola is at (0, 0). This is deduced because there are no linear terms \(x\) or \(y\) indicating shifts from the origin.

Step 3: Determine the Transverse and Conjugate Axes

The transversal axis is vertical because the \(y^2\) term is positive. From the equation \(\frac{y^2}{8} - \frac{x^2}{6} = 1\), we find \(a^2 = 8\) and \(b^2 = 6\).

Step 4: Calculate the Vertices

Using \(a^2 = 8\), calculate \(a = \sqrt{8} = 2\sqrt{2}\). The vertices are at \( (0, \pm 2\sqrt{2}) \).

Step 5: Calculate the Co-vertices

Using \(b^2 = 6\), calculate \(b = \sqrt{6}\). The co-vertices are at \( (\pm \sqrt{6}, 0) \).

Step 6: Identify the Asymptotes

The equations of the asymptotes for \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) are \( y = \pm \frac{a}{b}x \). Thus, \( y = \pm \frac{\sqrt{8}}{\sqrt{6}}x = \pm \frac{2\sqrt{3}}{3}x \).

Step 7: Graph the Hyperbola

Using a graphing device, plot the vertices \( (0, \pm 2\sqrt{2}) \) and co-vertices \( (\pm \sqrt{6}, 0) \). Draw the asymptotes \( y = \pm \frac{2\sqrt{3}}{3}x \), and sketch the branches of the hyperbola along the vertical transverse axis approaching these lines.

Key Concepts

Equation of Hyperbola

A hyperbola is defined by its standard equation, which can provide key information about its orientation, center, and axes. The equation we're looking at is given in the transformed standard form: \[ \frac{y^2}{8} - \frac{x^2}{6} = 1 \]This standard form means we are dealing with a hyperbola whose transverse axis is along the vertical direction because the positive term is \(y^2\). To transform any hyperbola equation into its standard form, follow these steps:

• Identify the terms associated with \(y^2\) and \(x^2\).
• Line the equation up to \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) or \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
• Divide all terms by the constant term to satisfy the right-hand side being 1.

This form helps quickly identify parameters like \(a^2\) and \(b^2\), which relate to the lengths of the axes. In this problem, you've found \(a^2 = 8\) and \(b^2 = 6\). These will be crucial for identifying the vertices and asymptotes. Remember, if \(y^2\) is the positive term, the transverse axis is vertical, otherwise it’s horizontal if \(x^2\) leads.

Vertices of Hyperbola

The vertices of a hyperbola are the points where its branches are closest to the center. These are crucial as they help in sketching the graph accurately. From the standard equation previously identified: \[ \frac{y^2}{8} - \frac{x^2}{6} = 1 \]We determine the vertices of the hyperbola by finding \(a\), the distance along the transverse axis. Here, \(a^2 = 8\), leading to \(a = \sqrt{8} = 2\sqrt{2}\).Hence, because the transverse axis is along the \(y\)-axis, the vertices are located at:

• \( (0, 2\sqrt{2}) \)
• \( (0, -2\sqrt{2}) \)

These points represent the upper and lower bounds of the hyperbola along its transverse axis in the vertical direction. Always remember:

• For a vertical transverse axis, vertices are above and below the center.
• For a horizontal transverse axis, vertices are to the left and right of the center.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are lines that the hyperbola approaches but never touches. They provide a framework for the branches of the hyperbola, helping in sketching it with accuracy.For the given equation:\[ \frac{y^2}{8} - \frac{x^2}{6} = 1 \]The asymptotes can be calculated using the formula:\[ y = \pm \frac{a}{b}x \]Here, \(a = \sqrt{8}\) and \(b = \sqrt{6}\), so the asymptotes equations are:\[ y = \pm \frac{\sqrt{8}}{\sqrt{6}}x = \pm \frac{2\sqrt{3}}{3}x \]These slanted lines form the boundary within which the hyperbola branches will be sketched. Key points to remember:

• For vertical transverse axis \((y^2\) term positive), asymptote formula is \(y = \pm \frac{a}{b}x\).
• For horizontal transverse axis \((x^2\) term positive), asymptote formula is \(y = \pm \frac{b}{a}x\).

These lines help in understanding the orientation and extent of the hyperbola branches. Always plot the asymptotes when sketching a hyperbola for guidance.