Problem 36 - Precalculus: Mathematics for Calculus

Question

Use a graphing device to graph the hyperbola. $$\frac{x^{2}}{100}-\frac{y^{2}}{64}=1$$

Step-by-Step Solution

Answer

Graph a hyperbola centered at (0,0) with vertices at (10,0) and (-10,0), with asymptotes at \( y=0.8x \) and \( y=-0.8x \).

Step 1: Identify the Equation Type

The equation \( \frac{x^2}{100} - \frac{y^2}{64} = 1 \) represents a hyperbola because it is in the standard form of a hyperbola with a horizontal transverse axis, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Observe that the term \( x^2 \) has a positive coefficient, confirming a horizontal orientation.

Step 2: Determine the Center

The hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is centered at \((h, k)\). In this equation, \( h = 0 \) and \( k = 0 \), thus the center is at the origin (0, 0).

Step 3: Calculate Vertices

For a horizontal hyperbola, vertices are \((h \pm a, k)\). Here, \( a = \sqrt{100} = 10 \). Thus, the vertices are at \((10, 0)\) and \((-10, 0)\).

Step 4: Calculate Foci

The foci of the hyperbola are calculated using \( c = \sqrt{a^2 + b^2} \). Given \( a = 10 \) and \( b = \sqrt{64} = 8 \), calculate \( c = \sqrt{10^2 + 8^2} = \sqrt{164} \). Thus, the foci are located at \((\sqrt{164}, 0)\) and \((-\sqrt{164}, 0)\).

Step 5: Plot Asymptotes

The equations of the asymptotes for a horizontal hyperbola are \( y = \pm \frac{b}{a}x \). Substituting \( b = 8 \) and \( a = 10 \), the equations become \( y = \pm \frac{8}{10}x = \pm 0.8x \).

Step 6: Graph the Hyperbola

Using a graphing device, plot the hyperbola with the vertices at \((10, 0)\) and \((-10, 0)\), the foci at \((\sqrt{164}, 0)\) and \((-\sqrt{164}, 0)\), and the asymptotes along \( y = 0.8x \) and \( y = -0.8x \). Adjust the viewing window to ensure all these points are visible.

Key Concepts

Standard Equation of a Hyperbola

The standard equation of a hyperbola is crucial in understanding its geometric behavior. For a hyperbola with a horizontal transverse axis, the equation is given by \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]This equation tells us that the hyperbola opens left and right. The value \( a \) represents the distance from the center to each vertex along the x-axis, while \( b \) refers to the "broadness" of the curve in the y direction, being the distance associated with the asymptotes.
Hyperbolas have a vertical transverse axis when \[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]which indicates they open upwards and downwards instead. Identifying whether the hyperbola is horizontal or vertical from its equation is straightforward:

• If the \( x^2 \) term is positive, it is horizontal.
• If the \( y^2 \) term is positive, it is vertical.

Knowing which variable is positive helps to set up expectations for how the hyperbola will graph.

Graphing Hyperbolas

Graphing hyperbolas might seem complex, but breaking it into steps makes it quite simple. First, identify the center, typically at point \((h, k)\), often the origin \((0, 0)\) in standard cases like this one.
The transverse axis determines the main stretch of the hyperbola.Find the vertices by using the equation's terms:

• For \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at \((h \pm a, k)\).
• For \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the vertices are at \((h, k \pm b)\).

Next, calculate and plot the foci. Use the formula \[ c = \sqrt{a^2 + b^2} \]locate the foci along the transverse axis at points \((h \pm c, k)\) or \((h, k \pm c)\), depending on orientation.
Asymptotes play a vital role in outlining the hyperbola.In the horizontal case, they are lines \(y = \pm \frac{b}{a}x\), and in the vertical, they are \(x = \pm \frac{a}{b}y\). Draw them as guides to shape your hyperbola curve correctly.

Hyperbola Vertices and Foci

Vertices and foci describe significant points on a hyperbola. These points help define its size and position in the coordinate plane. The vertices of a horizontal hyperbola like \( \frac{x^2}{100} - \frac{y^2}{64} = 1 \) are found at \((\pm a, 0)\). Calculating gives \( a = \sqrt{100} = 10 \), so vertices are \((10, 0)\) and \((-10, 0)\).
Foci of a hyperbola are determined by calculating \[ c = \sqrt{a^2 + b^2} \].For our example, with \( a = 10 \) and \( b = 8 \), we find \( c = \sqrt{164} \), placing the foci at \(( \sqrt{164}, 0)\) and \((-\sqrt{164}, 0)\).

• Vertices give us the "endpoints" of the hyperbola along the transverse axis.
• Foci indicate focal points within the hyperbola that aid in defining its geometric properties.

Remember, for graphing and calculating these points correctly,ensure to maintain their relative position to the center.