Problem 29 - 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. $$ y^{2}-x^{2}=8 $$

Step-by-Step Solution

Answer

Equation: \(\frac{y^2}{8} - \frac{x^2}{8} = 1\); Asymptotes: \(y = \pm x\); Foci: \((0, 4), (0, -4)\).

Step 1: Identify the Equation Type

The given equation is \(y^2 - x^2 = 8\), which is a form of a hyperbola equation. A standard form of a hyperbola equation is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) or \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, the equation resembles the form \(y^2 - x^2 = c\), indicating a vertical hyperbola.

Step 2: Convert to Standard Form

To convert \(y^2 - x^2 = 8\) into the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), divide the entire equation by 8:\\[ \frac{y^2}{8} - \frac{x^2}{8} = 1 \]\In this form, \(a^2 = 8\) and \(b^2 = 8\), making \(a = \sqrt{8} = 2\sqrt{2}\) and \(b = \sqrt{8} = 2\sqrt{2}\).

Step 3: Find the Asymptotes

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

Step 4: Find the Foci

The distance to each focus for a vertical hyperbola is given by \(c = \sqrt{a^2 + b^2}\). With \(a^2 = 8\) and \(b^2 = 8\), compute: \ \(c = \sqrt{8 + 8} = \sqrt{16} = 4\).\Thus, the foci are located at \((0, 4)\) and \((0, -4)\).

Step 5: Sketch the Hyperbola

1. Draw the asymptotes, \(y = x\) and \(y = -x\). These are diagonal lines passing through the origin.2. Mark the center of the hyperbola at the origin (0, 0).3. Since \(a = 2\sqrt{2}\), sketch the vertices at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).4. Plot the foci at \((0, 4)\) and \((0, -4)\).5. Draw the hyperbola opening vertically, with the center, vertices, asymptotes, and foci included in the sketch.

Key Concepts

Asymptotes in Hyperbolas

In a hyperbola, asymptotes are crucial as they guide the sketching of its shape. Asymptotes are straight lines that the hyperbola approaches but never touches.
They help to determine the hyperbola's orientation in the Cartesian plane.

• For a vertical hyperbola, given by the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the asymptotes are expressed as \(y = \pm \frac{a}{b}x\).
• For a horizontal hyperbola, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are \(y = \pm \frac{b}{a}x\).

By finding these lines, we can sketch hyperbolas more accurately and understand the graph's behavior at infinity.

Standard Form of a Hyperbola

Recognizing and converting to the standard form of a hyperbola allows us to find important characteristics such as vertices, foci, and asymptotes. The standard forms are:

• Vertical hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)
• Horizontal hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

These forms clearly show the denominators \(a^2\) and \(b^2\) as the squares of the distances from the center to the vertices along the transverse axis (major axis in the case of horizontal hyperbolas).
In our example, dividing by 8 transformed the equation \(y^2 - x^2 = 8\) into the standard form necessary to identify these key features.

Foci of a Hyperbola

The foci of a hyperbola are pivotal points in its geometry. Unlike ellipses, hyperbolas spread out around these focal points as they "open up." The foci are located along the transverse axis, and for hyperbolas:
- The distance from the center to a focus is \(c = \sqrt{a^2 + b^2}\).
- For vertical hyperbolas, the foci are at \((0, \pm c)\), and for horizontal hyperbolas, \((\pm c, 0)\).
In the given equation, \(c = \sqrt{16} = 4\), so the foci are positioned at \((0, 4)\) and \((0, -4)\). This placement helps in sketching the hyperbola accurately as they lie along the axis of symmetry.

Equation Transformation in Hyperbolas

Transforming an equation into its standard form is essential for understanding its properties and sketching it.
For hyperbolas, this typically involves:

• Rewriting or "normalizing" the equation to make one side equal to 1.
• Dividing the equation by the constant term that appears once both variables are isolated on one side.

This transformation uncovers \(a^2\) and \(b^2\), helping in determining the hyperbola's orientation and asymptotes.
In the example \(y^2 - x^2 = 8\), dividing by 8 gave \(\frac{y^2}{8} - \frac{x^2}{8} = 1\), putting it in the necessary form for analysis.