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

Divide by 8 to standardize, find asymptotes as \( y = x \) and \( y = -x \), and foci at \((0, 4)\), \((0, -4)\).

Step 1: Identify the Equation Type

Given the equation is \( y^2 - x^2 = 8 \), which is the equation of a hyperbola because it has a subtraction form between squared terms and one constant on the other side. A standard hyperbola equation has the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

Step 2: Rewrite in Standard Form

To convert the given equation into standard form, divide the entire equation by 8 to make the right side equal to 1: \[ \frac{y^2}{8} - \frac{x^2}{8} = 1 \]. Here, \(a^2 = 8\) and \(b^2 = 8\).

Step 3: Find the Asymptotes

For a hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), asymptotes are given by the equations \( y = \pm\frac{a}{b}x \). In our case, both \(a^2\) and \(b^2\) are equal, so \(a/b = 1\), leading to the asymptotes being \(y = x\) and \(y = -x\).

Step 4: Determine the Foci

For the hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the distance to the foci \(c\) is calculated using \(c^2 = a^2 + b^2\). Since \(a^2 = 8\) and \(b^2 = 8\), \(c^2 = 16\) giving \(c = 4\). The foci are situated along the y-axis at \( (0, 4) \) and \( (0, -4) \).

Step 5: Sketch the Hyperbola

Plot the hyperbola centered at the origin (0,0). Draw the asymptotes as lines \( y = x \) and \( y = -x \). Mark the foci at \( (0, 4) \) and \( (0, -4) \). The hyperbola should approach the asymptotes but never touch them, appearing like a stretched X shape.

Key Concepts

Understanding Hyperbola Asymptotes

When dealing with hyperbolas, asymptotes are essential lines that the curve approaches but never actually touches. These lines act as a guide on what path the arms of the hyperbola will follow as they extend infinitely. For the standard form of a hyperbola, which looks like \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), asymptotes can be easily derived. The equations for the asymptotes of a hyperbola are given by the formulas \( y = \pm \frac{a}{b}x \) when the transverse axis is vertical or \( x = \pm \frac{a}{b}y \) if the transverse axis is horizontal. Since the roles of \(a\) and \(b\) are the same in the transverse direction, the asymptotes are straightforward to compute:

• For a vertical transverse axis, the slopes of the asymptotes depend on \( \frac{a}{b} \).
• If \( a = b \), the slopes are \( \pm 1 \).

For example, the equation \( y^2 - x^2 = 8 \) results in the asymptotes \( y = x \) and \( y = -x \), depicting a cross of lines at the origin, which guides the hyperbola's arms.

Hyperbola Standard Form Made Easy

The concept of standard form is pivotal when working with hyperbolas because it simplifies interpretation and graphing. The standard form equation for hyperbolas is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) or \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), depending on the axis along which the hyperbola opens. This allows us to easily identify key attributes:

• Values of \(a^2\) and \(b^2\), which determine the size and orientation: here, both are equal to 8.
• The center of the hyperbola, in this case, at the origin (0,0).

To convert \( y^2 - x^2 = 8 \) into standard form, we divide every term by 8, giving us \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \). This setup lets us quickly read off necessary values for sketching and solving further problems related to asymptotes and foci.

Digging into the Foci of Hyperbolas

The foci of a hyperbola are points that lie along its primary axis, and they play a crucial role in defining the shape. For hyperbolas, the distance to the foci \(c\) from the center is calculated using the formula \( c^2 = a^2 + b^2 \). This relationship holds true for both horizontal and vertical hyperbolas. For our specific example, with \( a^2 = 8 \) and \( b^2 = 8 \), we find that \( c^2 = 16 \), leading to \( c = 4 \). Thus, the hyperbola's foci are located at points \( (0, 4) \) and \( (0, -4) \), assuming the hyperbola opens vertically along the y-axis. These foci are crucial for understanding the geometry of hyperbolas because:

• The hyperbola's shape exaggerates the distance between these focus points.
• The foci are always located within the arms of the hyperbola.

Marking these on a graph provides insight into how the hyperbola's curves bend around the center and extend towards infinity.