Problem 35 - Algebra and Trigonometry

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \(( \pm 1,0),\) asymptotes: \(y=\pm 5 x\)

Step-by-Step Solution

Answer

The equation is \(x^2 - \frac{y^2}{25} = 1\).

Step 1: Identify Hyperbola Type

Since the vertices are given as \(( \pm 1, 0)\), the hyperbola opens horizontally. For a horizontally oriented hyperbola, the standard form is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Given the vertices, the center \((h, k)\) is \((0, 0)\).

Step 2: Identify a Value

The vertices are at \((\pm 1,0)\), which means \(a = 1\) since the distance from the center to a vertex is \(a\). Therefore, we have \(a^2 = 1^2 = 1\).

Step 3: Identify Asymptotes Relation

The asymptotes are given by \(y=\pm 5x\). For a horizontal hyperbola, the slopes of the asymptotes \(\frac{b}{a}\) thus \(\frac{b}{1} = 5\), so \(b = 5\).

Step 4: Determine Equation

Now that we have \(a^2 = 1\) and \(b^2 = 25\), the equation of the hyperbola is \(\frac{x^2}{1} - \frac{y^2}{25} = 1\).

Step 5: Final Equation

The equation of the hyperbola is \(x^2 - \frac{y^2}{25} = 1\).

Key Concepts

Vertices of hyperbola

In a hyperbola, the vertices are the points where the hyperbola intersects its principal axis. This principal axis is the axis of symmetry of the hyperbola. In our specific case, the given vertices are \((\pm 1, 0)\). This tells us quite a few things about the hyperbola.

• Since the vertices have the form \((x,0)\), the hyperbola opens horizontally.
• The distance between these vertices is \(2a\), where \(a\) is the distance from the center to a vertex along the principal axis.

For this hyperbola, \(a = 1\). The information about the vertices is crucial in understanding the overall shape and orientation of the hyperbola. Knowing \(a\) helps us formulate parts of the hyperbola's equation.

Asymptotes of hyperbola

Asymptotes are the lines that the hyperbola approaches but never touches or crosses. They serve as important guides for the shape and direction of the branches of the hyperbola. In our case, the asymptotes are given by the equations \(y = \pm 5x\).

• These lines tell us that this is a horizontal hyperbola because the asymptotes have equations of the form \(y = \pm \frac{b}{a}x\).
• The slopes of the asymptotes, which are \(\pm 5\), indicate a high ratio between \(b\) and \(a\).

Using the relationship \(\frac{b}{a} = 5\), and knowing \(a = 1\), we find that \(b = 5\). Asymptotes help in sketching the hyperbola and are pivotal in determining its orientation and rate of opening.

Standard form of hyperbola

The standard form of the equation of a hyperbola is vital for understanding and deriving other information about the hyperbola. In this case, we are dealing with a horizontally opening hyperbola. Thus, the standard form is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]
- For our specific hyperbola, given the center \((h, k) = (0, 0)\), the equation simplifies to:\[\frac{x^2}{1} - \frac{y^2}{25} = 1\]- The terms in the standard form correspond to their respective orientations: the \(x\) term indicates it opens horizontally.
Understanding the standard form allows you to quickly identify characteristics like orientations, dimensions, and can provide you with the basis for sketching it accurately.

Center of hyperbola

The center of a hyperbola is a central point equidistant to the vertices and the open ends. For a hyperbola aligned with the coordinate axes, the center is denoted as \((h, k)\). In our task, the center of the hyperbola is at the origin \((0, 0)\).

• The center is important as it serves as a reference point for the entire structure of the hyperbola.
• It is used in both the vertex and asymptote equations.

Knowing the center allows one to accurately position the hyperbola within a graph and serves as the basis for calculating other parameters like \(a\), \(b\), and their squared equivalents in the hyperbola's equation.