Problem 38 - Algebra and Trigonometry

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) hyperbola passes through \((-5,9)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{y^2}{36} - \frac{x^2}{20} = 1\).

Step 1: Recognize the Orientation of the Hyperbola

Since the vertices are given as \((0, \pm 6)\), we know the hyperbola opens vertically. Thus, the general equation will be in the form of \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Here, \(a\) represents the distance from the center to the vertices.

Step 2: Determine the Value of \(a\)

The distance from the center \((0,0)\) to each vertex is \(6\). Hence, \(a = 6\), and \(a^2 = 36\).

Step 3: Solve for \(b^2\) Using the Point on the Hyperbola

The hyperbola passes through the point \((-5,9)\). Substitute \(x = -5\), \(y = 9\), and \(a^2 = 36\) into the equation \(\frac{y^2}{36} - \frac{x^2}{b^2} = 1\). This gives us: \(\frac{9^2}{36} - \frac{(-5)^2}{b^2} = 1\).

Step 4: Simplify the Equation and Solve for \(b^2\)

Calculate \(\frac{9^2}{36} = \frac{81}{36} = \frac{9}{4}\). This modifies the equation to: \(\frac{9}{4} - \frac{25}{b^2} = 1\). Simplifying further, \(\frac{9}{4} - 1 = \frac{25}{b^2}\). Thus, \(\frac{5}{4} = \frac{25}{b^2}\).

Step 5: Calculate \(b^2\)

Solve for \(b^2\) by multiplying both sides by \(b^2\) and then dividing by \(\frac{5}{4}\). This results in \(b^2 = \frac{25}{\frac{5}{4}} = 20\).

Step 6: Write the Final Equation of the Hyperbola

Substitute \(a^2 = 36\) and \(b^2 = 20\) back into the general form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). The final equation of the hyperbola is: \[\frac{y^2}{36} - \frac{x^2}{20} = 1\].

Key Concepts

Vertices

Understanding the concept of vertices is key to working with hyperbolas. In the context of hyperbolas, vertices are points on the hyperbola itself. They represent the closest distance from the center of the hyperbola to the curve. For a vertical hyperbola, these vertices are aligned along the y-axis. In our example, the vertices are at

• (0, 6)
• (0, -6)

This indicates that the hyperbola opens vertically. The vertex points are symmetric about the center of the hyperbola, which in this case is at the origin (0,0). Understanding the vertices helps determine the orientation and the equation of the hyperbola.

Equation of Hyperbola

The equation of a hyperbola is a mathematical representation that tells us about its shape and orientation. Hyperbolas can open either horizontally or vertically:

• Vertical hyperbolas use the form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
• Horizontal hyperbolas have the form: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

In our problem, the hyperbola is vertical due to the position of the vertices. For the vertical equation, "a" represents how far the vertices are from the center. Here, "a" is 6, so \(a^2 = 36\). We substitute known values for a and b into the formula to find the final equation: \[ \frac{y^2}{36} - \frac{x^2}{20} = 1 \] The values of "b" are usually found by using additional points the hyperbola passes through. This equation gives us a complete understanding of the hyperbola's proportions and placement.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never touches. They extend towards infinity in both directions. For a vertical hyperbola like ours, asymptotes give us valuable information about the hyperbola's shape, even though these lines aren't part of the hyperbola itself.The asymptotes' equations for a vertical hyperbola centered at (0,0) are:

• \[ y = \pm \frac{a}{b} x \]

Where "a" and "b" relate to the hyperbola's dimensions as noted in its equation. For our example, with \(a = 6\) and \(b = \sqrt{20}\), the asymptotes will be:\[ y = \pm \frac{6}{\sqrt{20}} x \]These asymptotes help visualize the open branches of the hyperbola as they extend towards infinity. Understanding how asymptotes work is essential for graphing hyperbolas accurately.

Conic Sections

Hyperbolas are a significant part of conic sections. Conic sections include several other shapes as well, including circles, ellipses, and parabolas. They are formed by intersecting a plane with a double-napped cone:

• A circle or ellipse is formed when the plane cuts the cone at an angle parallel or less than the slope angle of the cone.
• A parabola arises when the cut is parallel to the slope of the cone.
• Hyperbolas result when the plane cuts through both nappes of the cone.

By altering the angle and position of the intersecting plane, different conic sections are formed. Hyperbolas, in particular, consist of two separate curves or 'branches' that mirror each other. Understanding conic sections provides a framework for where and why hyperbolas fit within the study of geometry.