Problem 18 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 16 x^{2}-25 y^{2}-256 x-150 y+399=0 $$

Step-by-Step Solution

Answer

Center: (8, -3); Foci: \((8 \pm \sqrt{41}, -3)\); Vertices: \((3, -3)\) and \((13, -3)\); Asymptotes: \(y + 3 = \pm \frac{4}{5}(x - 8)\); Eccentricity: \(\frac{\sqrt{41}}{5}\).

Step 1: Write the Standard Form of the Hyperbola

The standard form of a hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\). Begin by rewriting the given equation to eventually match this form.

Step 2: Rearrange Terms

Move all terms involving \(x\) and \(y\) to the left side of the equation:\[16x^2 - 256x - 25y^2 - 150y = -399\]

Step 3: Complete the Square

First, complete the square for the \(x\) terms:\[16(x^2 - 16x) = 16((x - 8)^2 - 64) = 16(x-8)^2 - 1024\]Next, complete the square for the \(y\) terms:\[-25(y^2 + 6y) = -25((y + 3)^2 - 9) = -25(y+3)^2 + 225\]

Step 4: Simplify the Equation

Insert the complete squares back into the equation:\[16(x-8)^2 - 1024 - 25(y+3)^2 + 225 = -399\]Simplify the constant terms:\[16(x-8)^2 - 25(y+3)^2 = -399 + 1024 - 225= 400\]

Step 5: Divide by 400 to Standardize

Divide the entire equation by 400 to convert it to the standard form:\[\frac{16(x-8)^2}{400} - \frac{25(y+3)^2}{400} = 1\]Simplifying gives:\[\frac{(x-8)^2}{25} - \frac{(y+3)^2}{16} = 1\]

Step 6: Identify the Hyperbola Parameters

The center of the hyperbola is \((8, -3)\). The values \(a^2 = 25\) and \(b^2 = 16\) imply \(a = 5\) and \(b = 4\).

Step 7: Determine Foci and Vertices

For the hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are a distance of \(a\) along the \(x\)-axis from the center:Vertices: \((8 \pm 5, -3) = (13, -3)\) and \((3, -3)\).The distance \(c\), where \(c^2 = a^2 + b^2\), gives the foci:\[c = \sqrt{25 + 16} = \sqrt{41}\]Foci: \((8 \pm \sqrt{41}, -3)\).

Step 8: Calculate the Asymptotes

The slopes of the asymptotes for this hyperbola are \(\pm \frac{b}{a}\):Slopes: \(\pm \frac{4}{5}\).The equations of the asymptotes are:\[y + 3 = \pm \frac{4}{5}(x - 8)\]

Step 9: Determine the Eccentricity

The eccentricity \(e\) is given by \(e = \frac{c}{a}\):\[e = \frac{\sqrt{41}}{5}\].

Step 10: Graph the Hyperbola

Plot the center \((8, -3)\), plot the vertices at \((3, -3)\) and \((13, -3)\), and plot the foci at \((8\pm\sqrt{41}, -3)\). Draw the asymptotes with equations found and sketch the hyperbola respecting these lines.

Key Concepts

Center of Hyperbola

Every hyperbola has a central point, known as the center, which acts as the midpoint of the line segment joining the two vertices. This point is key because the entire hyperbola is defined around it. In this equation, after transforming it into standard form, we find that the hyperbola is centered at

• (8, -3)

To find this center, we completed the square for both the x and y terms and adjusted the equation until it fit the standard hyperbola form. Understanding where the center is in any hyperbola problem is crucial for graphing and other analyses.

Foci of Hyperbola

The foci of a hyperbola are two points inside each curve of the hyperbola. These points are vital because the definition of a hyperbola involves the absolute difference of distances from any point on the hyperbola to these foci being constant.
The foci give the hyperbola its unique shape, and in the standard position, they lie along the transverse axis of the hyperbola. For the given hyperbola, the foci can be calculated using the formula:

• \(c^2 = a^2 + b^2\)
• \(c = \sqrt{41}\)
• Points: \((8 \pm \sqrt{41}, -3)\)

This means the foci are located at approximately (14.4, -3) and (1.6, -3), providing the necessary reference for sketching the hyperbola.

Vertices of Hyperbola

The vertices of a hyperbola are points where the hyperbola intersects its transverse axis. These are the closest and farthest points to the center of the hyperbola when measured along the axis.
In the standard form \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\],

• Vertices occur at the points \((h \pm a, k)\).
• For our hyperbola:
• \(a = 5\)
• Vertices: \((8 \pm 5, -3)\) which translates to (13, -3) and (3, -3).

These are significant points that help define the scale and representation of the hyperbola on a graph.

Asymptotes of Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually reaches. They provide essential guidance for determining the orientation and "openness" of the hyperbola. Given the equation \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\],
The slopes of the asymptotes for a horizontal hyperbola are given by

• \(\pm \frac{b}{a}\) which, in this case, equals \(\pm \frac{4}{5}\).
• Their equations become:
• \(y + 3 = \pm \frac{4}{5}(x - 8)\).

These lines intersect at the hyperbola's center and guide the graphing process, ensuring that the arms of the hyperbola fall within these boundary lines.

Eccentricity of Hyperbola

The eccentricity of a hyperbola, denoted as \(e\), is a measure describing how stretched or elongated a hyperbola is.
It is calculated using the formula:

• \(e = \frac{c}{a}\)
• Where \(c = \sqrt{41}\) and \(a = 5\).
• So, \(e = \frac{\sqrt{41}}{5}\).

The value of \(e\) for hyperbolas will always be greater than 1. This is a distinguishing feature that sets hyperbolas apart from ellipses, which have an eccentricity less than 1. Understanding and calculating the eccentricity helps in assessing the shape and spread of the hyperbola.