Problem 12 - Algebra and Trigonometry

Question

Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$ \frac{(y-1)^{2}}{25}-(x+3)^{2}=1 $$

Step-by-Step Solution

Answer

Center: (-3, 1); Foci: (-3, 1±√26); Vertices: (-3, 6) and (-3, -4); Asymptotes: y-1=±5(x+3).

Step 1: Identify the Hyperbola Form

The given equation of the hyperbola is \( \frac{(y-1)^2}{25} - (x+3)^2 = 1 \). This is of the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), indicating a vertical hyperbola.

Step 2: Find the Center

The center of the hyperbola can be found from the form \((h, k)\). Here, \( h = -3 \) and \( k = 1 \). Therefore, the center is at \((-3, 1)\).

Step 3: Determine the Vertices

For a vertical hyperbola, vertices are at \((h, k \pm a)\). Given \(a^2 = 25\), so \(a = 5\). The vertices are at \((-3, 1+5)\) and \((-3, 1-5)\), which are \((-3, 6)\) and \((-3, -4)\).

Step 4: Calculate the Foci

For a vertical hyperbola, the foci are at \((h, k \pm c)\), where \(c^2 = a^2 + b^2\). \(b^2 = 1\), thus \(c^2 = 25 + 1 = 26\), leading to \(c = \sqrt{26}\). Therefore, the foci are at \((-3, 1 + \sqrt{26})\) and \((-3, 1 - \sqrt{26})\).

Step 5: Find the Asymptotes

The asymptotes of a vertical hyperbola are given by the equations \(y - k = \pm \frac{a}{b}(x - h)\). Here, \(a = 5\), \(b = 1\), and the center \((h, k)\) is \((-3, 1)\). Thus, the equations are \(y - 1 = \pm 5(x + 3)\).

Step 6: Sketch the Graph

Plot the center at \((-3, 1)\). Mark the vertices at \((-3, 6)\) and \((-3, -4)\). Plot the foci at approximately \((-3, 1 \pm 5.1)\) based on \(\sqrt{26} \approx 5.1\). Draw asymptotes using \(y - 1 = 5(x + 3)\) and \(y - 1 = -5(x + 3)\). Sketch the hyperbola opening vertically, away from the center, and approaching these asymptotes.

Key Concepts

Center of Hyperbola

The center of a hyperbola is a crucial point from which the structure of the hyperbola can be determined. For the equation \( \frac{(y-1)^2}{25} - (x+3)^2 = 1 \), the center can be found by identifying the values of \( h \) and \( k \) in the general form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Here, \( h = -3 \) and \( k = 1 \), making the center located at \((-3, 1)\). This point serves as the reference for plotting other key features of the hyperbola, like vertices and foci, contributing to an accurate graphical representation.

Foci of Hyperbola

The foci of a hyperbola are key points that help define its shape and orientation. For the given equation, since it is a vertical hyperbola, the foci are calculated using the formula \((h, k \pm c)\). First, we determine \(c\) using \(c^2 = a^2 + b^2\). Given \(a^2 = 25\) and \(b^2 = 1\), we have \(c^2 = 25 + 1 = 26\), so \(c = \sqrt{26}\). Therefore, the foci are positioned at \((-3, 1 + \sqrt{26})\) and \((-3, 1 - \sqrt{26})\). These points are located symmetrically above and below the center, further from it than the vertices.

Vertices of Hyperbola

Vertices are the points where each branch of the hyperbola makes its closest approach to the center. For our vertical hyperbola, the vertices are located using \((h, k \pm a)\). For this equation, \(a^2 = 25\) gives \(a = 5\). Thus, the vertices are \((-3, 1 + 5)\) and \((-3, 1 - 5)\) or \((-3, 6)\) and \((-3, -4)\).

• Closer to center: Vertices are closer to the center compared to the foci.
• Extends vertically: Confidence given it is a vertical hyperbola also ensures the vertices extend vertically.

Asymptotes of Hyperbola

Asymptotes provide important guidance in sketching the hyperbola since the branches of the hyperbola approach these lines as they extend towards infinity. For a vertical hyperbola, the asymptotes are calculated using the formula \(y - k = \pm \frac{a}{b}(x - h)\). In the equation given, we use \(a = 5\), \(b = 1\), and the center \((-3, 1)\). Therefore, the asymptotes are expressed as:

• \(y - 1 = 5(x + 3)\)
• \(y - 1 = -5(x + 3)\)

These lines help set the boundaries that the branches of the hyperbola will not cross.

Graph of Hyperbola

To graph a hyperbola, it is essential to plot all these components together. Start by marking the center at position \((-3, 1)\). From there, plot the vertices at \((-3, 6)\) and \((-3, -4)\) and the foci approximately at \((-3, 1 \pm 5.1)\) since \(\sqrt{26} \approx 5.1\). Use the asymptotes \(y - 1 = 5(x + 3)\) and \(y - 1 = -5(x + 3)\) as guidelines to sketch the hyperbola. Remember that the branches of a hyperbola are always drawn to open outwards from the center, with the vertices showing the closest points to the center and the foci directing the shape at large distances. This comprehensive layout ensures that each component is effectively used to visualize the hyperbola.