Problem 6 - 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. $$y^{2}-\frac{x^{2}}{15}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±1), Foci: (0, ±4), Asymptotes: y = ±x/√15.

Step 1: Identify the Type of Hyperbola

The standard form of a hyperbola equation is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) for a vertical hyperbola. Here, the given equation \( y^2 - \frac{x^2}{15} = 1 \) already matches this form, indicating it is a vertical hyperbola with \( a^2 = 1 \) and \( b^2 = 15 \).

Step 2: Calculate Vertices

For a vertical hyperbola, the vertices are located at \((0, \pm a)\). Given \( a^2 = 1 \), we find \( a = 1 \). Thus, the vertices are located at \((0, 1)\) and \((0, -1)\).

Step 3: Calculate Foci

The formula for calculating the distance to each focus for a hyperbola is \( c^2 = a^2 + b^2 \). Therefore, solve \( c^2 = 1 + 15 = 16 \), so \( c = 4 \). The foci are located at \((0, \pm 4)\).

Step 4: Find Equations of Asymptotes

For a vertical hyperbola, the equations of the asymptotes are given by \( y = \pm \frac{a}{b} x \). We have \( a = 1 \) and \( b = \sqrt{15} \), thus the equations are \( y = \pm \frac{1}{\sqrt{15}}x \).

Step 5: Sketch the Graph

Draw the hyperbola centered at the origin \((0,0)\), mark its vertices at \((0, 1)\) and \((0, -1)\), and its foci at \((0, 4)\) and \((0, -4)\). Draw dashed lines for the asymptotes \( y = \pm \frac{1}{\sqrt{15}}x \) to show behavior.

Key Concepts

Vertices of a Hyperbola

In hyperbola terms, vertices act like key markers along the path where the hyperbola changes direction. They tell us how wide or tall the shape is, according to its type. For a vertical hyperbola, the vertices sit along the y-axis. In the given equation, \(y^2 - \frac{x^2}{15} = 1\), we have a vertical hyperbola. The distance from the center of the hyperbola (in this case, the origin \((0,0)\)) to each vertex is represented by \(a\).

• The equation tells us that \(a^2 = 1\), so \(a = 1\).
• This means the vertices are located at \((0, 1)\) and \((0, -1)\).

It's helpful to picture the hyperbola as a stretched-out bow, with the vertices marking the peak points that help guide your sketch.

Foci of a Hyperbola

The foci of a hyperbola give defining characteristics, influencing how 'spread out' the hyperbola looks. These points lie farther from the center compared to the vertices, on the main axis which they share. To find the foci for a vertical hyperbola like ours, we use:

• \( c^2 = a^2 + b^2 \)
• From the equation: \( a^2 = 1 \) and \( b^2 = 15 \), so \( c^2 = 1 + 15 = 16 \).
• Thus, \( c = 4 \).
• Therefore, the foci are located at \((0, 4)\) and \((0, -4)\).

These foci don't appear explicitly on graphs, but they help shape the curves of the hyperbola, assisting in understanding its paths or branches.

Asymptotes of a Hyperbola

Asymptotes are imaginary guidelines that the hyperbola approaches but never touches. They give insight into the overall direction and steepness of the branches of the hyperbola. For a vertical hyperbola, our given hyperbola's asymptotes can be determined using the formula:

• \( y = \pm \frac{a}{b} x \)
• Here we calculate using \(a = 1\) and \(b = \sqrt{15}\).
• This gives us the asymptotes: \( y = \pm \frac{1}{\sqrt{15}}x \).

Drawing these as dashed lines lets you see how the hyperbola's branches open wide and guide the overall curve. It's as if these lines were the ceiling the hyperbola stretched towards but could never cross.