Problem 20 - Calculus Early Transcendentals

Question

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$\frac{y^{2}}{16}-\frac{x^{2}}{36}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±4); Foci: (0, ±√52); Asymptotes: y = ±(2/3)x.

Step 1: Identify the Standard Form

The given equation of the hyperbola is \( \frac{y^2}{16} - \frac{x^2}{36} = 1 \). This equation is in the standard form of a vertical hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), where \( a^2 = 16 \) and \( b^2 = 36 \).

Step 2: Calculate the Values of 'a' and 'b'

Solve for \( a \) and \( b \) by taking the square roots: \( a = \sqrt{16} = 4 \) and \( b = \sqrt{36} = 6 \).

Step 3: Determine the Vertices

The vertices of a vertical hyperbola are at \((0, \pm a)\). Substituting the value of \( a \), the vertices are \((0, 4)\) and \((0, -4)\).

Step 4: Calculate the Foci

The foci of a vertical hyperbola are located at \((0, \pm c)\), where \( c \) is calculated as \( c = \sqrt{a^2 + b^2} = \sqrt{16 + 36} = \sqrt{52} \). Hence, the foci are \((0, \sqrt{52})\) and \((0, -\sqrt{52})\).

Step 5: Find the Asymptotes

The equations of the asymptotes of a vertical hyperbola are \( y = \pm \frac{a}{b}x \). Substitute the known values: \( y = \pm \frac{4}{6}x = \pm \frac{2}{3}x \). So, the asymptotes are \( y = \frac{2}{3}x \) and \( y = -\frac{2}{3}x \).

Step 6: Sketch the Graph

To sketch the graph, plot the center at the origin \((0,0)\), then plot the vertices at \((0,4)\) and \((0,-4)\). Plot the foci at \((0, \sqrt{52})\) and \((0, -\sqrt{52})\), which are slightly farther from the center than the vertices. Draw the asymptotes through the origin with slopes \( \frac{2}{3} \) and \( -\frac{2}{3} \). Finally, sketch the hyperbola opening upwards and downwards approaching the asymptotes.

Key Concepts

Vertices

In the study of hyperbolas, one of the first essential components is understanding the vertices. The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. These are critical because they help define the shape and orientation of the hyperbola.
In a vertical hyperbola, like the one given in the exercise, the vertices are located along the y-axis. The equation used, \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \], shows us that the vertices appear at \((0, \pm a)\). Specifically, for your hyperbola, \(a = 4\) from \(a^2 = 16\), leading to vertices at \((0, 4)\) and \((0, -4)\).
This positioning tells us that the hyperbola opens upward and downward from the center located at the origin \((0,0)\). These vertices are not just points; they help guide the plot of the hyperbola and are pivotal in defining its branches.

Foci

In geometrical terms, the foci of a hyperbola are two points outside each curve from which the sum of distances to any point on the hyperbola is constant. The location of these foci is crucial in understanding the extent and nature of the hyperbola's shape.
For a vertical hyperbola, the foci are found at \((0, \pm c)\), where \(c\) is defined as: \[ c = \sqrt{a^2 + b^2} \]. In this exercise, substituting the values from the given equation results in:\[ c = \sqrt{16 + 36} = \sqrt{52} \].
Therefore, the foci are positioned at \((0, \sqrt{52})\) and \((0, -\sqrt{52})\). These points inform us about the "spread" of the hyperbola, as they lie farther from the center than the vertices do.
Visualizing these foci can be helpful when sketching, as they will ensure the hyperbola is drawn accurately upward and downward in relation to these defining points.

Asymptotes

The asymptotes of a hyperbola are straight lines that the hyperbola approaches but never actually meets. These lines are essential because they help delineate the overall direction in which the hyperbola opens.
For vertical hyperbolas, the asymptotes' equations are given by: \[ y = \pm \frac{a}{b}x \].In the current hyperbola, where \(a = 4\) and \(b = 6\), the asymptotes simplify to:\[ y = \pm \frac{4}{6}x \] or, reduced, \[ y = \pm \frac{2}{3}x \].
This provides linear guides showing the slopes at \(\frac{2}{3}\) and \(-\frac{2}{3}\) which the branches of the hyperbola will mimic as they extend infinitely away from the center. These invisible lines serve as crucial guides when sketching, ensuring the hyperbola draws consistently outwards, approaching, but never intersecting the asymptotes.