Problem 17 - Algebra and Trigonometry

Question

use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{y^{2}}{16}-\frac{x^{2}}{36}=1 $$

Step-by-Step Solution

Answer

The vertices are at (0,4) and (0,-4), the foci are at (0,7.21) and (0,-7.21), and the equations of the asymptotes are \( y= \frac{2}{3}x \) and \( y= -\frac{2}{3}x \)

Step 1: Identify a^2 and b^2

From the given equation, we can see that \( a^{2} = 16 \) and \( b^{2} = 36 \). Therefore, a = 4 and b = 6. The center of the hyperbola is at the origin (0,0).

Step 2: Find the vertices

The vertices occur at (0,±a), which is (0,±4). So the vertices are (0,4) and (0,-4).

Step 3: Determine the foci

Now we will find the foci. The foci are at (0,±c), where \( c = √(a^{2} + b^{2}) \) = √(16+36) = √52 ≈ ±7.21. So the foci are (0,7.21) and (0,-7.21).

Step 4: Determine the equations of the asymptotes

The equations of the asymptotes are of the form y = ±(a/b)x. Here, this will be \( y = ±\frac{4}{6}x = ±\frac{2}{3}x \). So the asymptotes are \( y= \frac{2}{3}x \) and \( y= -\frac{2}{3}x \).

Key Concepts

Understanding Vertices in a Hyperbola

The vertices of a hyperbola are essential points that provide valuable information about its shape and orientation. When dealing with the hyperbola \( \frac{y^{2}}{16}-\frac{x^{2}}{36}=1 \), we start by identifying our center, which is at the origin (0,0). Now, let's see how we find the vertices:

• Determine \( a^2 \): From the equation, \( a^2 = 16 \), thus \( a = 4 \).
• Locate the vertices: The vertices occur at the points (0,±a), so for this hyperbola, the vertices are at (0,4) and (0,-4).

These vertices define the most extended points along the transverse axis, which, in this case, is the y-axis. Remember, they help set the boundaries of the hyperbola's branches and can be used as a guide when sketching the graph.

Finding the Foci of the Hyperbola

The foci are two special points located along the transverse axis of a hyperbola. These points are not only crucial for its definition but also play a role in its geometric properties. To determine the foci, follow these steps:

• Calculate \( c \): Use the formula \( c = \sqrt{a^2 + b^2} \). Here, \( a^2 = 16 \) and \( b^2 = 36 \),making \( c = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \).
• Position of the foci: Since the hyperbola opens vertically, the foci are at \( (0,±c) \), which gives us (0,7.21) and (0,-7.21).

These points indicate where the hyperbola bends and determine its shape further. Understanding the foci's location helps to comprehend the spatial structure and curvature of the graph.

Exploring Asymptotes in Hyperbolas

Asymptotes are lines that the hyperbola approaches but never intersects. They provide a framework outlining how the hyperbola extends into infinity. To find the asymptotes for our hyperbola, we need to derive their equations:

• General Form: Asymptotes for hyperbolas centered at the origin are given by \( y = \pm \frac{a}{b} x \).
• Substitute Values: With \( a = 4 \) and \( b = 6 \), the equations become \( y = \pm \frac{4}{6} x = \pm \frac{2}{3} x \).
• Equation of asymptotes: For our hyperbola, the equations of the asymptotes are \( y = \frac{2}{3}x \) and \( y = -\frac{2}{3}x \).

These lines help us to visualize the steepness and direction of the hyperbola's branches. By sketching these lines, we ensure the symmetry and alignment of the graph, and they act as a guide in depicting an accurate representation.

Graphing Hyperbolas Using Vertices, Foci, and Asymptotes

To successfully sketch a hyperbola, combining the information from vertices, foci, and asymptotes is key. Here's a straightforward method to graph the hyperbola \( \frac{y^{2}}{16}-\frac{x^{2}}{36}=1 \):

• Draw the vertex points: Mark the vertices at (0,4) and (0,-4).
• Plot the foci: Position the foci at (0,7.21) and (0,-7.21). Ensure these are marked farther along the y-axis.
• Sketch the asymptotes: Draw the lines \( y = \frac{2}{3}x \) and \( y = -\frac{2}{3}x \) through the center.
• Outline the hyperbola: Using the vertices and the asymptotes as guides, sketch the curve approaching the asymptotes but not crossing them. The branches should be symmetrical about the y-axis.

This method provides a clear pathway to illustrate the hyperbola's shape, ensuring that the drawn graph aligns with the calculated coordinates.