Problem 11 - Precalculus: Mathematics for Calculus

Question

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$\frac{y^{2}}{36}-\frac{x^{2}}{4}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±6), Foci: (0, ±2√10), Asymptotes: y = ±3x, Transverse axis: 12.

Step 1: Identify the Hyperbola's Standard Form

The equation given is \( \frac{y^2}{36} - \frac{x^2}{4} = 1 \). This matches the standard form of a vertical hyperbola, \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), where \( a^2 = 36 \) and \( b^2 = 4 \). Consequently, \( a = 6 \) and \( b = 2 \).

Step 2: Find the Vertices

The vertices of a vertical hyperbola are located at \((0, \pm a)\). Since \( a = 6 \), the vertices are at \((0, 6)\) and \((0, -6)\).

Step 3: Determine the Foci

To find the foci, use the formula \( c^2 = a^2 + b^2 \). Here, \( c^2 = 36 + 4 = 40 \). Thus, \( c = \sqrt{40} = 2\sqrt{10} \). The foci, being vertically oriented, are \((0, \pm 2\sqrt{10})\).

Step 4: Calculate the Asymptotes

For a vertical hyperbola, the equations of the asymptotes are \( y = \pm \frac{a}{b}x \). Substituting the values, the asymptotes are given by \( y = \pm \frac{6}{2}x = \pm 3x \).

Step 5: Determine the Length of the Transverse Axis

The transverse axis is the distance between the vertices, calculated as \( 2a = 2 \times 6 = 12 \).

Step 6: Sketch the Graph of the Hyperbola

Plot the vertices \((0, 6)\) and \((0, -6)\), the foci \((0, 2\sqrt{10})\) and \((0, -2\sqrt{10})\), and draw the asymptotes \( y = 3x \) and \( y = -3x \). The hyperbola opens upwards and downwards along the y-axis with these guiding lines.

Key Concepts

Vertices

In the study of hyperbolas, vertices are key points that help define the shape. For a standard vertical hyperbola like \(\frac{y^2}{36} - \frac{x^2}{4} = 1\), the vertices are located along the y-axis. This is due to the fact that the formula involves \((0, \pm a)\), where \(a\) is derived from \(a^2 = 36\), giving us \(a = 6\). These points mark the positions where the hyperbola intersects its transverse axis.
Therefore, in this particular equation, the vertices are at:

• \((0, 6)\)
• \((0, -6)\)

This means the vertex points are 6 units up and down from the center \((0,0)\). Visualizing them helps in sketching and understanding the structure of the hyperbola.

Foci

The foci of a hyperbola are important as they are points from which the distances to any point on the hyperbola maintain a constant difference. For the equation \(\frac{y^2}{36} - \frac{x^2}{4} = 1\), the foci are calculated using \(c^2 = a^2 + b^2\).
Here, we calculate:
\[c^2 = 36 + 4 = 40\]
Thus,\(c = \sqrt{40} = 2\sqrt{10}\).
The foci, being vertically placed, are at:

• \((0, 2\sqrt{10})\)
• \((0, -2\sqrt{10})\)

These lie further away from the center than the vertices and are crucial in defining the hyperbola's open structure. Remember, the foci for a vertical hyperbola always lie on the y-axis.

Asymptotes

Asymptotes in a hyperbola are crucial lines that the hyperbola approaches but never actually intersects. For vertical hyperbolas, asymptotes help guide the shape.Let's use the equation for asymptotes: \(y = \pm \frac{a}{b}x\). With \(a = 6\) and \(b = 2\), we find:
\[y = \pm \frac{6}{2}x = \pm 3x\]
These asymptotes describe two diagonal lines that cross through the origin (0,0) and extend diagonally. They create a sort of boundary that the hyperbola's arms will approach but never touch. Asymptotes help in anticipating the hyperbola's stretch and direction in a graph, guiding the sketching process.

Transverse Axis

The transverse axis is a fundamental part of a hyperbola's structure. For a vertical hyperbola, this axis is aligned along the y-axis, connecting the two vertices directly. It essentially describes how wide or tight the opening of the hyperbola appears.
In our example equation \(\frac{y^2}{36} - \frac{x^2}{4} = 1\), the transverse axis is determined by measuring the distance between the vertices. Since the vertices are at \((0, 6)\) and \((0, -6)\), the total length of the transverse axis is found using:
\[2a = 2 \times 6 = 12\]
This means that there's 12 units total length directly through and along the y-axis between the vertices. Understanding the transverse axis helps in visualizing and constructing the graph of the hyperbola, showing how wide it stretches vertically.

Graph of Hyperbola

Drawing the graph of a hyperbola involves putting all the previous concepts together. Our equation \(\frac{y^2}{36} - \frac{x^2}{4} = 1\) showcases a hyperbola opening along the y-axis, a characteristic of vertical hyperbolas.
Start by plotting the vertices \((0, 6)\) and \((0, -6)\). Then, draw the asymptotes \(y = 3x\) and \(y = -3x\), ensuring they pass through the origin and extend outwards, creating a guiding framework for the hyperbola.
The foci \((0, 2\sqrt{10})\) and \((0, -2\sqrt{10})\) lie further out along the vertical line, showing the hyperbola's direction and ensuring the graph aligns within the boundaries set by the asymptotes.
By connecting these points and lines, you'll achieve a curve that gets infinitely closer to the asymptotes as it moves away from the center, providing a clear depiction of the hyperbola's "opening" nature along its transverse axis. Remember, the arms of the hyperbola never touch the asymptotes, always curving gracefully without crossing these bounds.