Problem 5 - Algebra and Trigonometry with Analytic Geometry

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$x^{2}-\frac{y^{2}}{24}=1$$

Step-by-Step Solution

Answer

Vertices: \((1, 0)\), \((-1, 0)\); Foci: \((5, 0)\), \((-5, 0)\); Asymptotes: \(y = 2\sqrt{6}x\), \(y = -2\sqrt{6}x\).

Step 1: Identify the Hyperbola Form

The given hyperbola equation is \( x^2 - \frac{y^2}{24} = 1 \). This equation is of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 1 \) and \( b^2 = 24 \).

Step 2: Calculate the Vertices

The vertices of a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are located at \( (\pm a, 0) \). So, \( a = \sqrt{1} = 1 \). Therefore, the vertices are located at \( (1, 0) \) and \( (-1, 0) \).

Step 3: Determine the Foci

The foci of the hyperbola are found using the formula \( c = \sqrt{a^2 + b^2} \). Calculate \( c = \sqrt{1 + 24} = \sqrt{25} = 5 \). Thus, the foci are at \( (\pm 5, 0) \).

Step 4: Find the Equations of Asymptotes

The asymptotes of a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by \( y = \pm \frac{b}{a}x \). Here, \( \frac{b}{a} = \frac{\sqrt{24}}{1} = \sqrt{24} = 2\sqrt{6} \). Therefore, the equations of the asymptotes are \( y = 2\sqrt{6}x \) and \( y = -2\sqrt{6}x \).

Step 5: Graph the Hyperbola

To sketch the graph, start by plotting the center at the origin. Mark the vertices at \( (1, 0) \) and \( (-1, 0) \), and the foci at \( (5, 0) \) and \( (-5, 0) \). Draw the asymptotes, which are lines through the origin with slopes \( 2\sqrt{6} \) and \( -2\sqrt{6} \). The hyperbola will open horizontally around these asymptotes.

Key Concepts

Vertices

The vertices of a hyperbola are key points that help define its shape and orientation. For a hyperbola centered at the origin, expressed in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are positioned along the x-axis at \((\pm a, 0)\). In our example, \( a^2 = 1 \) implies \( a = \sqrt{1} = 1 \). Consequently, the vertices are located at \((1, 0)\) and \((-1, 0)\). These points outline the extent of the hyperbola along the x-axis, determining how wide it appears. To visualize, think of the vertices as anchor points on a rubber band stretched horizontally. As you move away from the origin, the hyperbola curves outwards, beginning at these vertices and extending towards infinity, constrained by the asymptotes. This clear identification of vertices is crucial when sketching or analyzing the graph of the hyperbola.

Foci

The foci are critical elements in the geometry of a hyperbola, helping to comprehend its defining properties. Essentially, for any point on a hyperbola, the difference of its distances to the foci is constant. For our hyperbola, given the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are calculated using the formula \( c = \sqrt{a^2 + b^2} \).

• Here, \( a^2 = 1 \) and \( b^2 = 24 \), leading to \( c = \sqrt{1 + 24} = 5 \).
• This results in the foci being located at \((\pm 5, 0)\).

The foci play a pivotal role by identifying points where the hyperbola appears to `focus` around. Picture them as drill holes around which two cones of light expand outward, intersecting along the horizontal axis. Recognizing the position of the foci helps fully depict the structural anatomy of the hyperbola, enabling more comprehensive analysis and an accurate sketch.

Asymptotes

Asymptotes in the context of a hyperbola are guiding lines that the hyperbola approaches but never actually meets. They effectively "frame" the curved branches, providing a template for the general spread of the hyperbola as it extends towards infinite points. For a hyperbola with the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations for the asymptotes are described by \( y = \pm \frac{b}{a}x \).

• In the given equation, \( b = \sqrt{24} = 2\sqrt{6} \), and \( a = 1 \).
• This implies our asymptotes are \( y = 2\sqrt{6}x \) and \( y = -2\sqrt{6}x \).

When graphing, these asymptotes resemble criss-crossing lines, ideal as boundaries the hyperbola edges towards but never crosses. Think of asymptotes as invisible fences guiding the hyperbola’s arms in a clear direction without physical restraint. These lines, therefore, provide a critical framework to accurately represent how the hyperbola behaves, ensuring its portrayal is symmetrical and balanced.