Problem 31 - Thomas Calculus

Question

Exercises \(27-34\) give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. $$ 8 x^{2}-2 y^{2}=16 $$

Step-by-Step Solution

Answer

The hyperbola's standard form is \(\frac{x^2}{2} - \frac{y^2}{8} = 1\) with asymptotes \(y = \pm 2x\). The foci are at \((\pm \sqrt{10}, 0)\).

Step 1: Arrange the Equation

The given equation is \(8x^2 - 2y^2 = 16\). Begin by rearranging the equation so that it equates to 1. Divide every term by 16:\[\frac{8x^2}{16} - \frac{2y^2}{16} = \frac{16}{16}\]Simplifying gives:\[\frac{x^2}{2} - \frac{y^2}{8} = 1\]

Step 2: Identify the Type of Hyperbola

The equation \(\frac{x^2}{2} - \frac{y^2}{8} = 1\) is in the hyperbola standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 2\) and \(b^2 = 8\). This suggests a horizontally oriented hyperbola.

Step 3: Find the Asymptotes

For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are given by the equations \(y = \pm \frac{b}{a}x\).Calculate \(\frac{b}{a}\):- \(a = \sqrt{2}\) and \(b = \sqrt{8} = 2\sqrt{2}\)- \(\frac{b}{a} = \frac{2\sqrt{2}}{\sqrt{2}} = 2\)Thus, the asymptotes are:\(y = \pm 2x\).

Step 4: Determine the Foci

The foci of a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\).Calculate \(c\):- \(c = \sqrt{2 + 8} = \sqrt{10}\)Thus, the foci are at \((\pm \sqrt{10}, 0)\).

Step 5: Sketch the Hyperbola

To sketch the hyperbola, plot the center at the origin (0,0). The vertices are located \(\sqrt{2} \approx 1.41\) units away from the origin along the x-axis. Draw the asymptotes, which intersect at the center and follow the lines \(y = 2x\) and \(y = -2x\). Finally, plot the foci at \((\pm \sqrt{10}, 0)\). The hyperbola branches will approach the asymptotes but never touch them. The result is two curves extending left and right along the x-axis.

Key Concepts

Standard Form Equations for Hyperbolas

Hyperbolas, like other conic sections, have a standard form that makes it easier to identify their characteristics. To find the standard form, you need to rearrange the equation of the hyperbola so that it equates to 1. This step helps in simplifying the expressions and allows for easy comparison with the known form of hyperbolas. Let's consider the original equation: \(8x^2 - 2y^2 = 16\). By dividing each term by 16, we obtain: \[\frac{x^2}{2} - \frac{y^2}{8} = 1\] This places the equation in the standard hyperbolic form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), signifying a horizontally oriented hyperbola. Here, \(a^2 = 2\) and \(b^2 = 8\). The values \(a\) and \(b\) are essential for determining other properties of the hyperbola, like its asymptotes and foci. Thus, the standard form is not only a blueprint but a gateway to unlocking more about the hyperbola's geometry.

Understanding Hyperbola Asymptotes

Asymptotes are lines that the branches of a hyperbola approach but never actually intersect. They act as boundary markers that give direction to the hyperbola's curvature. For a hyperbola described by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes have the equations \(y = \pm \frac{b}{a}x\). These equations are derived from the characteristics of the conic section's geometry. In our specific case with \(\frac{x^2}{2} - \frac{y^2}{8} = 1\), we determine the slopes of the asymptotes:

• \(a = \sqrt{2}\)
• \(b = 2\sqrt{2}\)
• \(\frac{b}{a} = 2\)

Therefore, the asymptotes become \(y = 2x\) and \(y = -2x\). These lines intersect at the hyperbola's center, guiding its arms across the plane. Although they never touch the hyperbola, asymptotes are crucial for accurately graphing the hyperbola, providing a skeletal frame that correctly shapes the graph's form.

Locating the Foci of Hyperbolas

The foci are two special points in the hyperbola that hold great importance in its formation and definition. For a hyperbola, the foci are positioned along the major axis and are used to describe the mathematical property that differentiates a hyperbola from other conics. In standard form, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci are at \((\pm c, 0)\) for a horizontally oriented hyperbola.
To find the distance to the foci, \(c\), we use the formula \(c = \sqrt{a^2 + b^2}\). For our example, we calculate:

• \(a^2 = 2\)
• \(b^2 = 8\)

Hence, \(c = \sqrt{2 + 8} = \sqrt{10}\). This places the foci at \((\pm \sqrt{10}, 0)\). The positions of the foci directly influence the shape of the hyperbola, as they determine how "spread apart" the branches are. When graphing, these points ensure that each curve of the hyperbola maintains its proper distance and direction.