Problem 15 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 8(x+4)^{2}-5(y-7)^{2}+40=0 $$

Step-by-Step Solution

Answer

Center: (-4, 7); Vertices: \((-1.764, 7)\), \((-6.236, 7)\); Foci: \((-0.394, 7)\), \((-7.606, 7)\); Eccentricity: 1.613; Asymptotes: \(y = 7 \pm 1.264(x+4)\).

Step 1: Rewrite the Equation in Standard Form

To rewrite the equation of the hyperbola in standard form, we first need to isolate the terms involving x and y. Start by moving the constant to the other side: \[ 8(x+4)^2 - 5(y-7)^2 = -40 \] Next, divide every term by -40 to simplify the equation:\[ \frac{(x+4)^2}{5} - \frac{(y-7)^2}{8} = 1 \] Now it resembles the standard form of a hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).

Step 2: Identify the Center

The center (\( h, k \)) of the hyperbola is given by the transformations \((x-h)\) and \((y-k)\): \( h = -4 \) and \( k = 7 \). So, the center of the hyperbola is at \((-4, 7)\).

Step 3: Find the Vertices

The vertices are along the x-axis (since the terms involving x are positive and leading): \[ Vertices = (h \pm a, k) = (-4 \pm \sqrt{5}, 7) \]Calculate \( a \): \( \sqrt{5} \) is approximately 2.236. Thus, the vertices are: \[ (-4 + \sqrt{5}, 7) \approx (-1.764, 7) \] \[ (-4 - \sqrt{5}, 7) \approx (-6.236, 7) \]

Step 4: Find the Foci

The foci are calculated using the formula \( c = \sqrt{a^2 + b^2} \): \( b^2 = 8 \) and \( a^2 = 5 \)\[ c = \sqrt{5 + 8} = \sqrt{13} \approx 3.606 \]The foci positions are: \[ (-4 \pm \sqrt{13}, 7) \approx (-0.394, 7) \text{ and } (-7.606, 7) \]

Step 5: Calculate the Asymptotes

The equations of the asymptotes for the hyperbola are given by: \( y = k \pm \frac{b}{a}(x - h) \)By substituting \( b = \sqrt{8} \approx 2.828 \) and \( a = \sqrt{5} \approx 2.236 \), the asymptotic lines become:\( y = 7 \pm \frac{2.828}{2.236}(x + 4) \)This simplifies to:\( y = 7 \pm 1.264(x + 4) \)

Step 6: Determine the Eccentricity

The eccentricity \( e \) is found using the relation \( e = \frac{c}{a} \): \( e = \frac{\sqrt{13}}{\sqrt{5}} \approx \frac{3.606}{2.236} \approx 1.613 \)

Step 7: Graph the Hyperbola

To graph the hyperbola, plot the center at (-4, 7), the vertices at approximate locations (-1.764, 7) and (-6.236, 7), and the foci at approximately (-0.394, 7) and (-7.606, 7). Use the asymptotes \( y = 7 \pm 1.264(x + 4) \) as guidelines for drawing the hyperbola's branches opening horizontally.

Key Concepts

Eccentricity

The concept of eccentricity is crucial when studying hyperbolas. It measures how much a conic section deviates from being circular. For hyperbolas, the eccentricity is always greater than 1. This means hyperbolas are stretched further apart compared to an ellipse, which has an eccentricity between 0 and 1.
In the given exercise, the eccentricity \( e \) was calculated using the formula:\[ e = \frac{c}{a} \]Where \( c \) is the distance from the center to each focus and \( a \) is the distance from the center to a vertex.

• For this hyperbola, \( c = \sqrt{13} \approx 3.606 \)
• \( a = \sqrt{5} \approx 2.236 \)

Substituting these values into the formula gives:\[ e \approx 1.613 \]This higher value of \( e \) confirms that our hyperbola is much more elongated than a circle or an ellipse.

Vertices of a Hyperbola

Vertices play a key role in understanding the shape and orientation of a hyperbola. They are the points where the hyperbola intersects its principal axis. Every hyperbola has two vertices.
For the specific hyperbola given in the exercise, the vertices are calculated along the x-axis because it opens horizontally:\[ (h \pm a, k) \]Where \( h = -4 \) and \( k = 7 \) are the coordinates of the center, and \( a = \sqrt{5} \). Thus, you find the vertices:

• The vertex to the right is: \((-4 + \sqrt{5}, 7)\approx (-1.764, 7)\)
• The vertex to the left is: \((-4 - \sqrt{5}, 7)\approx (-6.236, 7)\)

They help define the basic "width" and direction of the hyperbola, with both vertices and the center lying on a straight horizontal line.

Foci of a Hyperbola

The foci of a hyperbola are fundamental in its defining properties. Foci are located inside each branch of the hyperbola and help in determining the shape of the hyperbola.
Using the relation:\[ c = \sqrt{a^2 + b^2} \]We find the distance \( c \) from the center to each focus. In our case:

• \( a^2 = 5 \)
• \( b^2 = 8 \)

This results in \( c \approx \sqrt{13} \approx 3.606 \). Therefore, the foci for this hyperbola are located at:

• \((-4 + \sqrt{13}, 7) \approx (-0.394, 7)\)
• \((-4 - \sqrt{13}, 7) \approx (-7.606, 7)\)

These points further away from the center than the vertices, confirm the hyperbolic nature of this conic.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never touches. They provide a framework helping to draw and understand the shape of the hyperbola.
For a hyperbola of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the equations of the asymptotes are given by:\[ y = k \pm \frac{b}{a}(x - h) \]In this exercise:

• \( a = \sqrt{5} \approx 2.236 \)
• \( b = \sqrt{8} \approx 2.828 \)

After plugging in the values, the equations become:

• \( y = 7 \pm \frac{2.828}{2.236}(x + 4) \)

Which simplifies to:

• \( y = 7 \pm 1.264(x + 4) \)

These asymptotic lines cross at the center of the hyperbola and indicate the direction of the branches of the hyperbola.