Problem 11 - Algebra 2

Question

Find the foci of each hyperbola. Then draw the graph. $$ \frac{y^{2}}{49}-\frac{x^{2}}{64}=1 $$

Step-by-Step Solution

Answer

The foci of the given hyperbola are located at \((0, sqrt{113})\) and \((0, - sqrt{113})\). The graph of the hyperbola has the center at the origin (0,0), vertices at (0, ±7), and the foci located as calculated above.

Step 1: Identifying values of a and b from the equation

From the given equation, we can see that \(a^2 = 49\) and \(b^2 = 64\). Therefore, a = 7 and b = 8.

Step 2: Calculate the coordinates of foci

The foci are located at \((0,\pm c)\) where c is determined by the equation \(c = sqrt{a^2 + b^2}\). Substitute a = 7 and b = 8 into this equation to compute the value of c, the distance from the center to each focus. So, \(c = sqrt{49 + 64} = sqrt{113}\). Thus, the coordinates of the foci are \((0, sqrt{113})\) and \((0, - sqrt{113})\).

Step 3: Drawing the graph

To draw the hyperbola graph, it's important to identify the key components from the given equation. The center of the hyperbola is at the origin (0,0). The vertices are (0, ±a) i.e., (0,±7). Draw the vertical transverse axis as a vertical line going through the center. Then draw the horizontal conjugate axis as a horizontal line through the center. Then sketch the graph with vertices and asymptotes.

Key Concepts

Foci of Hyperbolas

In a hyperbola, the foci are two special points located along the principal axis that help define its shape and properties. The equation for our hyperbola is \[\frac{y^{2}}{49} - \frac{x^{2}}{64} = 1\]This indicates that it is a vertical hyperbola because the \(y^2\) term is positive. To find the foci, we need to calculate the value of \(c\), which represents the distance from the center to each focus. The formula is \(c = \sqrt{a^2 + b^2}\).

• Calculate \(a\) and \(b\): We have \(a^2 = 49\), so \(a = 7\). Similarly, \(b^2 = 64\), giving \(b = 8\).
• Find \(c\): Using the formula, \(c = \sqrt{49 + 64} = \sqrt{113}\).
• Foci Coordinates: The foci for a vertical hyperbola are located at \((0, \pm c)\), which in this case are \((0, \sqrt{113})\) and \((0, -\sqrt{113})\).

The foci are crucial because a hyperbola is defined as the set of points such that the absolute difference of the distances to the two foci is constant.

Graphing Hyperbolas

Graphing a hyperbola involves plotting its shape accurately based on its specific equation. Here, our hyperbola's equation \(\frac{y^{2}}{49} - \frac{x^{2}}{64} = 1\) shows it is centered at the origin with a vertical transverse axis.

• Identify the Center: The center is at (0,0) since the equation is not shifted from the origin.
• Determine the Vertices: For vertical hyperbolas, vertices are on the \(y\)-axis. Since \(a = 7\), vertices are (0,7) and (0,-7).
• Draw the Transverse Axis: This is the main line through the vertices, vertically in our case.
• Sketch the Asymptotes: Asymptotes intersect at the center and have slopes given by \(\pm \frac{a}{b}\). They guide the hyperbola's opening.

The hyperbola will open vertically, and its shape will mirror around the asymptotes, illustrating the curve's wide spread at the extremities.

Equations of Hyperbolas

Equations of hyperbolas come in different forms, often based on their orientation. Understanding these forms is key to analyzing and graphing hyperbolas.

• Standard Form: The standard form for a hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for vertical hyperbolas and \(\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1\) for horizontal hyperbolas.
• Identify Parameters: In our equation, \(a^2 = 49\) and \(b^2 = 64\), showing \(a = 7\) and \(b = 8\). These help plot the vertices and asymptotes.

The equation reveals crucial properties, such as the direction the hyperbola opens—vertically for this equation—as well as distances to vertices and foci.

Conjugate Axis

The conjugate axis is a less intuitive part of hyperbolas. It doesn’t intersect the curve itself but provides important geometric context. It is crucial for correctly sketching the hyperbola.

• Definition: The conjugate axis is perpendicular to the transverse axis at the center. It acts as a kind of 'spacer'.
• Length: The length of the conjugate axis is \(2b\), which in our instance is \(16\).
• Graph Appearance: Though it isn't part of the hyperbola curve, it's drawn to define the rectangle where asymptotes are framed.

The conjugate axis aids in defining the hyperbola’s geometry, making it crucial for graph completion and understanding overall symmetry. Its interaction with the transverse axis and asymptotes helps create the asymptotic box, guiding the hyperbola's arms.