Problem 7 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ y^{2}-5 x^{2}=20 $$

Step-by-Step Solution

Answer

The center is (0,0), vertices are (0,±2√5), foci are (0,±2√6), asymptotes are y=±√5x, and eccentricity is √(6/5).

Step 1: Write the Equation in Standard Form

The given equation is \( y^{2} - 5x^{2} = 20 \). To write this equation in the standard form of a hyperbola, divide every term by 20, the number on the right side. This gives us \( \frac{y^{2}}{20} - \frac{5x^{2}}{20} = 1 \), simplifying further, \( \frac{y^{2}}{20} - \frac{x^{2}}{4} = 1 \).

Step 2: Identify the Hyperbola Type and Locate the Center

From the standard form \( \frac{y^{2}}{20} - \frac{x^{2}}{4} = 1 \), we recognize that the hyperbola's transverse axis is vertical since the \( y^{2} \) term is positive. The center of the hyperbola is at \((h, k)\), which is \((0, 0)\) for this equation.

Step 3: Find Vertices of the Hyperbola

For a vertical hyperbola \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the vertices are at \( (h, k \pm a) \). Here, \( a^{2} = 20 \) so \( a = \sqrt{20} = 2\sqrt{5} \). Therefore, the vertices are located at \((0, \pm 2\sqrt{5})\).

Step 4: Calculate the Foci

The foci of a vertical hyperbola \((centered~on~h,~,k)\) are given by \( (h, k \pm c) \), where \( c = \sqrt{a^{2} + b^{2}} \). Here, \( b^{2} = 4 \), so \( c = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6} \). Therefore, the foci are \((0, \pm 2\sqrt{6})\).

Step 5: Determine Asymptotes

The asymptotes of a vertical hyperbola are given by the lines \( y = \pm \frac{a}{b}x \). Here, \( a = \sqrt{20} = 2\sqrt{5} \) and \( b = 2 \). Therefore, the equations of the asymptotes are \( y = \pm \frac{2\sqrt{5}}{2}x = \pm \sqrt{5}x \).

Step 6: Calculate the Eccentricity

The eccentricity \( e \) of a hyperbola is found using \( e = \frac{c}{a} \). Here, \( c = 2\sqrt{6} \) and \( a = 2\sqrt{5} \), so \( e = \frac{2\sqrt{6}}{2\sqrt{5}} = \frac{\sqrt{6}}{\sqrt{5}} \approx 1.095 \).

Step 7: Graph the Hyperbola

Draw the coordinate axes and plot the center at \( (0, 0) \). Plot the vertices at \( (0, \pm 2\sqrt{5}) \) and the foci at \((0, \pm 2\sqrt{6})\). Draw dashed lines for the asymptotes \( y = \pm \sqrt{5}x \) to guide the shape. Sketch the hyperbola opening upwards and downwards in line with these guidelines.

Key Concepts

Standard Form of Hyperbola

To analyze a hyperbola thoroughly, it's crucial to express its equation in the standard form. This form helps in identifying its center and axis orientation, among other characteristics. The standard form of a hyperbola can look different based on its orientation:

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

For our given equation \( y^{2} - 5x^{2} = 20 \), we divide by 20 to derive \( \frac{y^{2}}{20} - \frac{x^{2}}{4} = 1 \), indicating a vertical transverse axis. Recognizing whether the \( y^{2} \) or the \( x^{2} \) term comes first tells us the direction in which the hyperbola opens.

Vertices of Hyperbola

Vertices serve as vital reference points determining the extent of the hyperbola along its transverse axis. For a vertical hyperbola in the form \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the vertices are located at \( (h, k \pm a) \).

From our equation, with \( a^2 = 20 \), we find \( a = \sqrt{20} = 2\sqrt{5} \). This means the vertices for our hyperbola are

• \((0, +2\sqrt{5})\)
• \((0, -2\sqrt{5})\)

These points help in sketching the hyperbola as they represent the corners, giving a visual sense of its height.

Foci of Hyperbola

The foci are fixed points that determine the shape and the extent of the hyperbola. For a hyperbola, the formula to find the foci is dependent on both \( a \) and \( b \). Notably, for a vertical hyperbola such as ours, the foci are positioned at \( (h, k \pm c) \), where \( c = \sqrt{a^{2} + b^{2}} \).

Given \( a^2 = 20 \) and \( b^2 = 4 \), we find \( c = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6} \). Thus, the foci are positioned at

• \((0, +2\sqrt{6})\)
• \((0, -2\sqrt{6})\)

These points are farther from the center than the vertices, further out along the transverse axis.

Asymptotes of Hyperbola

Asymptotes provide a guiding framework for the hyperbola's general shape, especially at its outermost reaches. For vertical hyperbolas, asymptotes are expressed as \( y = \pm \frac{a}{b}x \).

With our hyperbola, using \( a = 2\sqrt{5} \) and \( b = 2 \), the asymptotes become the lines

• \( y = +\sqrt{5}x \)
• \( y = -\sqrt{5}x \)

These straight lines intersect the origin and diverge such that the hyperbola hugs closer to these lines as one moves further from the center.

Eccentricity of Hyperbola

The eccentricity, denoted as \( e \), measures how much a conic section deviates from being circular, and helps in assessing the "strength" or "sharpness" of the hyperbola's curves. For hyperbolas, it's given by the formula \( e = \frac{c}{a} \).

With our numbers, \( c = 2\sqrt{6} \) and \( a = 2\sqrt{5} \), thus the eccentricity is calculated as

• \( e = \frac{\sqrt{6}}{\sqrt{5}} \approx 1.095 \)

An eccentricity greater than 1 confirms the existence of a hyperbola, with higher values indicating a more "open" shape.