Problem 13 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 25(x-3)^{2}-5(y-1)^{2}=125 $$

Step-by-Step Solution

Answer

Center: (3, 1); Foci: (3±√30, 1); Vertices: (3±√5, 1); Asymptotes: y=1±√5(x-3); Eccentricity: √6.

Step 1: Write the standard form of the hyperbola equation

The given equation is \( 25(x-3)^2 - 5(y-1)^2 = 125 \). Divide every term by 125 to set it to 1: \( \frac{25(x-3)^2}{125} - \frac{5(y-1)^2}{125} = 1 \). Simplify this to get \( \frac{(x-3)^2}{5} - \frac{(y-1)^2}{25} = 1 \).

Step 2: Identify the center of the hyperbola

In the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center is \((h, k)\). Here, \(h = 3\) and \(k = 1\), so the center is \((3, 1)\).

Step 3: Determine the vertices

For a hyperbola in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the vertices are \((h \pm a, k)\). Here, \(a = \sqrt{5}\), so the vertices are at \((3 \pm \sqrt{5}, 1)\).

Step 4: Calculate the foci

The distance from the center to each focus is \(c\) and is given by \(c^2 = a^2 + b^2\). Here, \(a^2 = 5\) and \(b^2 = 25\), so \(c = \sqrt{30}\). The foci are at \((3 \pm \sqrt{30}, 1)\).

Step 5: Determine the asymptotes

For a hyperbola of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), asymptotes are given by \(y = k \pm \frac{b}{a}(x-h)\). Here, \(\frac{b}{a} = \frac{5}{\sqrt{5}} = \sqrt{5}\), so the asymptotes are \(y = 1 \pm \sqrt{5}(x-3)\).

Step 6: Find the eccentricity

The eccentricity of a hyperbola is given by \(e = \frac{c}{a}\). Using \(c = \sqrt{30}\) and \(a = \sqrt{5}\), the eccentricity is \(e = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}\).

Step 7: Graph the hyperbola

Using the center \((3, 1)\), vertices \((3 \pm \sqrt{5}, 1)\), foci \((3 \pm \sqrt{30}, 1)\), and the asymptotes, draw the hyperbola opening horizontally on a coordinate plane.

Key Concepts

Conic Sections

Conic sections are shapes that you get when a plane intersects a cone. There are four primary types of conic sections: circle, ellipse, parabola, and hyperbola. Each has unique properties and equations that define its shape. Conic sections are pervasive across various fields such as astronomy, physics, and engineering. They help us understand orbits, satellite dishes, and even reflectors in cars. To identify a particular conic section, observe the general form of their equation:

• Circular: Equations like \(x^2 + y^2 = r^2\), where every point is equidistant from a central point.
• Elliptic: Equations like \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), with two axes of different lengths.
• Parabolic: Equations like \(y^2 = 4ax\), representing a curve opening in one direction.
• Hyperbolic: Equations like \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), describing two distinct curves.

Eccentricity of a Hyperbola

Eccentricity is a measure that describes how stretched or "un-round" a conic section is. For hyperbolas, the eccentricity is always greater than 1. This is because hyperbolas are more elongated or stretched than ellipses. The formula to calculate the eccentricity of a hyperbola involves the distance between the foci and the vertices:

• Given a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the distance to the foci is \(c = \sqrt{a^2 + b^2}\).
• The eccentricity \(e\) is then \(e = \frac{c}{a}\).

This number gives you an insight into the shape of the hyperbola. An eccentricity close to 1 means the branches of the hyperbola are closer together. Larger numbers mean they are more open.

Graphing Hyperbolas

Graphing a hyperbola can seem daunting, but by following systematic steps, you can easily form the characteristic hyperbolic shape on a graph. Here's how to tackle this:

• Start with the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(-\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
• Identify the center \((h, k)\), which is pivotal for plotting other elements.
• Locate the vertices, positioned along the transverse axis, and plot them relative to the center using \(a\).
• Determine the foci positions using \(c\), which are further away compared to the vertices.
• Draw lines called asymptotes that pass through the center. They act as guiding lines for the hyperbola's arms, helping you sketch the curve approaching but never touching these lines.
• Finally, sketch the two separate branches of the hyperbola, curving away from the center and closely following the asymptotes.

Asymptotes of Hyperbolas

Asymptotes are straight lines that the hyperbola approaches but never touches. They provide a guideline for the curve of each branch of the hyperbola. For hyperbolas, the asymptotes are determined by the slopes related to the hyperbola's axes.

• In the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the equations of the asymptotes are \(y = k \pm \frac{b}{a}(x-h)\).
• This involves calculating a slope of \(\frac{b}{a}\) that dictates the steepness of the asymptotes.

Understanding asymptotes is crucial because they sketch an invisible boundary that helps describe the hyperbola's overall shape. They form an "X" pattern that the hyperbola hugs as it extends out further on the graph.