Problem 21 - Calculus Early Transcendentals

Question

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$y^{2}-x^{2}=4$$

Step-by-Step Solution

Answer

Vertices: (0, ±2). Foci: (0, ±2√2). Asymptotes: y = ±x.

Step 1: Identify the Hyperbola Standard Equation Form

First, recognize that the given equation is of the form \(y^2 - x^2 = 4\), which matches the standard form of a hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Here, \(a^2 = 4\) and \(b^2 = 4\).

Step 2: Find the Vertices

For hyperbolas in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\). Since \(a^2 = 4\), \(a = 2\). Thus the vertices are at \((0, 2)\) and \((0, -2)\).

Step 3: Calculate the Foci

The foci of a hyperbola are located at \((0, \pm c)\), where \(c\) is found using the equation \(c^2 = a^2 + b^2\). Here, \(c^2 = 4 + 4 = 8\), so \(c = \sqrt{8} = 2\sqrt{2}\). The foci are at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).

Step 4: Determine the Asymptotes

The equations of the asymptotes for hyperbolas centered at the origin with the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) are \(y = \pm \frac{a}{b}x\). Since \(a = 2\) and \(b = 2\), the asymptotes are \(y = \pm x\).

Step 5: Sketch the Graph

Draw the transverse axis on the y-axis with vertices at \((0, 2)\) and \((0, -2)\). Plot the foci at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\). Draw the asymptotes \(y = x\) and \(y = -x\). The hyperbola opens up and down along the y-axis approaching these asymptotes.

Key Concepts

Vertices of Hyperbola

Understanding the vertices of a hyperbola is crucial in analyzing its shape and orientation. In general, the vertices of a hyperbola serve as the points where the hyperbola intersects its transverse axis. For the equation given, the hyperbola is represented by the standard form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]where the vertices are found at \( (0, \pm a) \). Here, \( a^2 = 4 \), so \( a = 2 \). This means our vertices are precisely at \( (0, 2) \) and \( (0, -2) \). These points are significant as they mark the beginning and the end of the central section of the hyperbola, which is the widest portion of the curve. The specific orientation of these vertices along the y-axis indicates that the hyperbola opens upwards and downwards rather than sideways. This trait makes hyperbolas similar to vertical parabolas. Understanding this can guide you in accurately sketching the curve's shape on a graph.

Foci of Hyperbola

The foci of a hyperbola are special points located along its transverse axis, contributing to defining the hyperbola's shape. Foci are calculated using the formula \( c^2 = a^2 + b^2 \),where \( c \) denotes the distance from the center to each focus. In our exercise, \( a^2 = 4 \) and \( b^2 = 4 \),resulting in \( c^2 = 4 + 4 = 8 \).Solving this gives \( c = \sqrt{8} = 2\sqrt{2} \). Thus, the foci are located at \( (0, 2\sqrt{2}) \) and \( (0, -2\sqrt{2}) \). These points sit outside of the vertices, further along the y-axis, and essentially act as invisible "gravitational centers" that dictate the hyperbola's 'pull' and shape. When sketching the hyperbola, plot these points accurately, as they will determine how tightly or loosely the hyperbola curves around its transverse axis.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches; they provide the framework upon which the hyperbola "rests". They can be visualized as the support beams that establish the overall spread of the hyperbola.For a hyperbola centered at the origin, given by the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \),the asymptotes have the equations \( y = \pm \frac{a}{b}x \). Knowing \( a = 2 \) and \( b = 2 \) means that \( \frac{a}{b} = 1 \). Thus, the asymptotes here are the lines \( y = x \) and \( y = -x \).When drawing the hyperbola, extend the lines \( y = x \) and \( y = -x \) through the origin. These lines help visually guide where the two branches of the hyperbola open and spread. The hyperbola will stretch and get infinitely closer to these lines as it moves away from its center, providing a direction for its curves. As such, asymptotes are integral to accurately portraying the behavior and shape of a hyperbola.