Problem 20 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ \frac{(y-4)^{2}}{16}-\frac{(x+2)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Vertices: \((-2, 8), (-2, 0)\), Foci: \((-2, 9), (-2, -1)\), Asymptotes: \(y = 4 \pm \frac{4}{3}(x + 2)\).

Step 1: Identify center, vertices, and orientation

The equation of a hyperbola in the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) suggests the hyperbola is vertical. Here, \( h = -2 \), \( k = 4 \), \( a^2 = 16 \) implying \( a = 4 \), and \( b^2 = 9 \) implying \( b = 3 \). The center of the hyperbola is at \((-2, 4)\). The hyperbola opens upwards and downwards.

Step 2: Calculate vertices

For a vertical hyperbola, vertices are given by \((h, k \pm a)\). Using \( h = -2 \), \( k = 4 \), and \( a = 4 \): Vertices: \((-2, 4+4) = (-2, 8) \) and \((-2, 4-4) = (-2, 0) \).

Step 3: Calculate foci coordinates

For a vertical hyperbola, foci are given by \((h, k \pm c)\), where \( c^2 = a^2 + b^2 \). Here, \( c^2 = 16 + 9 = 25 \) leading to \( c = 5 \). Thus, foci: \((-2, 4+5) = (-2, 9)\) and \((-2, 4-5) = (-2, -1)\).

Step 4: Determine equations of asymptotes

Asymptotes of a vertical hyperbola are given by \( y = k \pm \frac{a}{b}(x-h) \). With \( a = 4 \), \( b = 3 \), \( h = -2 \), \( k = 4 \), the equations are:\( y = 4 \pm \frac{4}{3}(x + 2) \), leading to:1. \( y = 4 + \frac{4}{3}(x + 2) \)2. \( y = 4 - \frac{4}{3}(x + 2) \).

Key Concepts

Vertices

Vertices are key points on a hyperbola. They represent the closest points of each branch of the hyperbola to the center. Vertices of a vertical hyperbola are found by adding and subtracting the value of \( a \) from the \( y \)-coordinate of the center. In this problem, the center is at \((-2, 4)\). Thus, the vertices are calculated as follows:
\[ (h, k \pm a) = (-2, 4 \pm 4) \]
This gives us the vertices at \((-2, 8)\) and \((-2, 0)\). These vertices lie directly above and below the center.

Foci

The foci of a hyperbola are crucial because they define the shape and orientation. For a vertical hyperbola, like the one given, the foci are located along the \( y \)-axis from the center. The distance from the center to each focus is \( c \), where \( c\) is calculated as the square root of \( a^2 + b^2 \). In this case, we find:
\[ c^2 = 16 + 9 = 25 \]
Therefore, \( c = 5 \). The foci are then positioned at:
\[ (h, k \pm c) = (-2, 4 \pm 5) \]
which results in the coordinates \((-2, 9)\) and \((-2, -1)\). These points help to better plot and understand the hyperbola's extent.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never touches. They provide a framework within which the hyperbola sits. For a vertical hyperbola, the equations of asymptotes are given by
\( y = k \pm \frac{a}{b}(x-h) \). Using the values from our problem \( a = 4 \), \( b = 3 \), \( h = -2 \), and \( k = 4 \), we derive:
\[ y = 4 \pm \frac{4}{3}(x + 2) \]
Expressed further, these become:
\[ y = 4 + \frac{4}{3}(x + 2) \]
\[ y = 4 - \frac{4}{3}(x + 2) \]
Asymptotes guide the direction of the hyperbola's branches and give us an idea of how the hyperbola opens.

Vertical Hyperbola

A vertical hyperbola, such as the one in this problem, opens up and down. This is identified by the structure of the equation \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). In a vertical hyperbola, the \( y \)-term is positive, and the structure of the equation determines that the branches of the hyperbola extend pointing upwards and downwards.
This orientation affects the placement of vertices, foci, and the alignment of asymptotes.
It is essential for correctly graphing the horizontal and vertical expansion of the hyperbola.

Center of Hyperbola

The center of a hyperbola is the midpoint between its vertices. It serves as the reference point for sketching other features of the hyperbola like vertices, foci, and asymptotes. In a hyperbola equation, the center is given by \((h, k)\).
For the given problem, the center can be extracted directly from the equation:
\( (h, k) = (-2, 4) \).
From this center position, all other calculations like the distances to the vertices and foci are derived. Understanding the center's position is critical for plotting the hyperbola accurately.