Problem 83 - Intermediate Algebra

Question

$$ \text { Graph: } 16 x^{2}-25 y^{2}=1 $$

Step-by-Step Solution

Answer

Graph a hyperbola centered at (0, 0), with vertices (1/4, 0) and (-1/4, 0), and asymptotes through y=±(5/4)x.

Step 1: Identifying the Conic Section Type

The given equation is \(16x^2 - 25y^2 = 1\). Notice that it is similar to the standard form of a hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). We identify this as a hyperbola because the signs of the \(x^2\) and \(y^2\) terms are opposite.

Step 2: Rewrite in Standard Form

The given equation is \(16x^2 - 25y^2 = 1\). To convert this to its standard hyperbola form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), divide through by 1: \(\frac{16x^2}{1} - \frac{25y^2}{1} = 1\), then rewrite it as \(\frac{x^2}{(1/4)^2} - \frac{y^2}{(1/5)^2} = 1\).

Step 3: Identify the Parameters

In \(\frac{x^2}{(1/4)^2} - \frac{y^2}{(1/5)^2} = 1\), \(a^2 = \frac{1}{16}\) and \(b^2 = \frac{1}{25}\), so \(a = \frac{1}{4}\) and \(b = \frac{1}{5}\). Thus, the center is at \((0, 0)\), as the equation lacks an \(x-h\) or \(y-k\) transformation.

Step 4: Graph the Hyperbola

The center of the hyperbola is at \((0, 0)\). From the center, move \(a = \frac{1}{4}\) units left and right along the x-axis to get the vertices at \((\frac{1}{4}, 0)\) and \((-\frac{1}{4}, 0)\). Move \(b = \frac{1}{5}\) units up and down along the y-axis to mark the co-vertices but remember they are not part of the graph; they aid in sketching the rectangular box. Finally, draw asymptotes through \(y = \pm\frac{5}{4}x\) using these vertices.

Key Concepts

Hyperbola

In the world of conic sections, a hyperbola is a fascinating shape that arises from slicing a double cone with a plane. It consists of two separate curves called branches. The defining characteristic of a hyperbola is having two squared terms with opposite signs when written in standard form. For example, in the equation \( 16x^2 - 25y^2 = 1 \), the positive and negative signs of \( x^2 \) and \( y^2 \) identify this equation as representing a hyperbola.A hyperbola features certain attributes:

• Center: The point that serves as the midpoint of the line segment that joins the vertices of the hyperbola. For hyperbolas with the center at the origin, the transverse axis is the x-axis if \( a > b \) and the y-axis if \( b > a \).
• Vertices: They lie along the transverse axis, equidistant from the center.
• Asymptotes: These are straight lines that the hyperbola tends to approach as it proceeds further away from the center.

With these properties, hyperbolas are not just useful in mathematics but also in physics and engineering, modeling phenomena like the paths of comets or the design of reflectors.

Graphing Hyperbolas

Graphing a hyperbola might seem challenging, but by following clear steps, it becomes manageable. Let's build on the equation \( \frac{x^2}{(1/4)^2} - \frac{y^2}{(1/5)^2} = 1 \).First, determine the center of the hyperbola. In this case, since there is no \( x-h \) or \( y-k \) in the equation, the center is at the origin \((0, 0)\). Next, identify the distances along each axis by finding the values of \(a\) and \(b\). Here, \(a = \frac{1}{4}\) and \(b = \frac{1}{5}\).Steps to Graph:

• Vertices: From the center, move \(a\) units left and right along the x-axis. This results in vertices at \((\frac{1}{4}, 0)\) and \((-\frac{1}{4}, 0)\).
• Co-vertices: Although not part of the hyperbola, move \(b\) units up and down along the y-axis to get helpful reference points \((0, \frac{1}{5})\) and \((0, -\frac{1}{5})\).
• Asymptotes: Draw lines through these points that pass through the origin using the slope \(\pm \frac{b}{a}\) or \(\pm \frac{5}{4}x\).

Asymptotes guide the path of each branch of the hyperbola. By sketching these components, visualize the hyperbola's open arms extending towards infinity.

Standard Form of Conics

The standard form of conic sections is crucial for understanding their geometric properties and graphing them accurately. Each conic type has a unique signature based on its equation. For hyperbolas, the standard form is typically \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).Converting to standard form simplifies:

• Identification: Standard form makes it easy to identify the type of conic section: circle, ellipse, hyperbola, or parabola. The different signs in the squared terms highlight a hyperbola.
• Parameters: In standard form, \(a\) and \(b\) are parameters that provide distances from the center to key points like vertices and enable the drawing of the asymptotes.
• Symmetry: These equations reveal symmetry, helping to place the hyperbola accurately on the graph.

Learning to rewrite equations into this form aids not just in graphing but also in deriving deeper insights into relations between variables, enhancing understanding across broader mathematical contexts.