Problem 26 - Intermediate Algebra

Question

Sketch the graph of each equation. \(4 y^{2}-25 x^{2}=100\)

Step-by-Step Solution

Answer

The graph is a vertical hyperbola centered at (0,0) with vertices at (0,5) and (0,-5), and asymptotes \(y=\frac{5}{2}x\) and \(y=-\frac{5}{2}x\).

Step 1: Identify the Conic Section

The given equation is \(4y^2 - 25x^2 = 100\). Notice that both \(y^2\) and \(x^2\) appear, and they have opposite signs. This indicates that the equation represents a hyperbola.

Step 2: Rewrite the Equation

Divide the entire equation by 100 to convert it to the standard form for a hyperbola. This gives us \(\frac{4y^2}{100} - \frac{25x^2}{100} = 1\). Simplifying, this becomes \(\frac{y^2}{25} - \frac{x^2}{4} = 1\).

Step 3: Determine the Center and Orientation

The hyperbola \(\frac{y^2}{25} - \frac{x^2}{4} = 1\) is centered at the origin \((0,0)\) because there are no terms involving \(x\) and \(y\). The \(y^2\) term comes first, indicating that the transverse (main) axis is vertical.

Step 4: Identify the Vertices and Co-vertices

For the hyperbola \(\frac{y^2}{25} - \frac{x^2}{4} = 1\), the vertices are determined by \(\sqrt{25}\) along the \(y\)-axis, giving points \((0, 5)\) and \((0, -5)\). The co-vertices are \((\sqrt{4}, 0)\), i.e., \((2, 0)\) and \((-2, 0)\) along the \(x\)-axis.

Step 5: Calculate and Draw the Asymptotes

The equations for the asymptotes of the hyperbola are given by \(\frac{y}{5} = \pm \frac{x}{2}\). Therefore, the asymptotes are \(y = \frac{5}{2}x\) and \(y = -\frac{5}{2}x\). These lines pass through the center and help sketch the hyperbola.

Step 6: Sketch the Hyperbola

Now sketch the graph: Plot the center at \((0,0)\), the vertices at \((0,5)\) and \((0,-5)\), and the co-vertices at \((2,0)\) and \((-2,0)\). Draw the asymptotes \(y = \frac{5}{2}x\) and \(y = -\frac{5}{2}x\). Then sketch the hyperbola, opening upwards and downwards, approaching but never touching the asymptotes.

Key Concepts

Conic Sections

Conic sections are a fascinating and fundamental topic in geometry that includes circles, ellipses, parabolas, and hyperbolas. Each of these shapes can be derived from slicing a cone at different angles. In the context of the given exercise, the hyperbola is our focus. A hyperbola is formed when a plane intersects both nappes (the two opposite conical surfaces) of a double cone.
A key feature that distinguishes a hyperbola from other conic sections is its pair of open, mirror-image curves. Unlike ellipses and parabolas, hyperbolas have this unique form because their equations include both squared terms with opposite signs. In this case, the original equation, \(4y^2 - 25x^2 = 100\), reveals a hyperbola due to the presence of both \(y^2\) and \(x^2\) having opposite signs. Hyperbolas are highly valuable in mathematical modeling, such as in astronomy and physics, where they describe natural phenomena like the paths of comets.

Graphing Equations

When graphing conic sections such as a hyperbola, it is crucial to first convert the equation into its standard form. For a hyperbola centered at the origin, the standard form is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) or \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). These forms inform us about the orientation and dimensions of the hyperbola.
In our example, the equation \(\frac{y^2}{25} - \frac{x^2}{4} = 1\) aligns with the first standard form, where \(a^2 = 25\) and \(b^2 = 4\). This means the hyperbola opens upwards and downwards since the \(y^2\) term comes first. Plotting such graphs involves more than just plotting points; it requires understanding and interpreting the equation's structure. An effective strategy is to map out the center, vertices, co-vertices, and asymptotic lines to guide the sketching of the hyperbola.

Asymptotes

Asymptotes are critical lines for understanding the behavior of a hyperbola as they indicate the direction in which the curves of the hyperbola approach infinitely. For hyperbolas centered at the origin, the equations for the asymptotes are derived from the equation's standard form. Asymptotes help provide a framework within which the hyperbola is sketched.
From \(\frac{y^2}{25} - \frac{x^2}{4} = 1\), the asymptotes can be determined from the equation \(\frac{y}{a} = \pm \frac{x}{b}\). Substituting the values, we get \(\frac{y}{5} = \pm \frac{x}{2}\), resulting in asymptotes \(y = \frac{5}{2}x\) and \(y = -\frac{5}{2}x\). These lines pass through the center of the hyperbola and remain a useful guide to ensure the curves of the hyperbola are properly sketched. The asymptotes assure that the curves hug these lines but never actually touch them, showcasing the infinite nature of a hyperbola.

Vertices and Co-vertices

Vertices and co-vertices are essential attributes of hyperbolas for defining their size and orientation. The vertices are the closest points to the center on the hyperbola's main axis, while the co-vertices lie on the axis perpendicular to it. They are vital in sketching and understanding the shape of the hyperbola.
In the given equation, we identify the vertices using \(\sqrt{a^2}\) along the \(y\)-axis, providing vertices at \((0,5)\) and \((0,-5)\). Co-vertices are found using \(\sqrt{b^2}\) along the \(x\)-axis, resulting in points \((2,0)\) and \((-2,0)\). This information defines the "box" that helps draw the hyperbola, positioned at the intersection of the vertices and co-vertices. By marking these points on a graph, one can draw a rectangle that helps determine the shape and proximity of the hyperbola's curves to the asymptotes, serving as a visual reference for accurate graphing.