Problem 12 - Precalculus: Mathematical for Calculus

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: \((-3, 1)\), Vertices: \((-3, 6)\), \((-3, -4)\), Foci: \((-3, 1 \pm \sqrt{26})\), Asymptotes: \(y = 5x + 16\), \(y = -5x - 14\).

Step 1: Recognize the Standard Form

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

Step 2: Identify the Center

From the equation \(\frac{(y-1)^{2}}{25}-(x+3)^{2}=1\), we compare to \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) and determine \(h = -3\) and \(k = 1\). Hence, the center of the hyperbola is \((-3, 1)\).

Step 3: Determine the Vertices

For a vertical hyperbola, the vertices are located \(a\) units above and below the center. Here, \(a^2 = 25\), thus \(a = 5\). The vertices are at \((-3, 1 + 5) = (-3, 6)\) and \((-3, 1 - 5) = (-3, -4)\).

Step 4: Locate the Foci

The foci are \(c\) units from the center along the transverse axis, where \(c = \sqrt{a^2 + b^2}\). Here, \(b^2 = 1\) and \(a^2 = 25\). Calculating \(c = \sqrt{25 + 1} = \sqrt{26}\). The foci are at \((-3, 1 + \sqrt{26})\) and \((-3, 1 - \sqrt{26})\).

Step 5: Find the Asymptotes

The asymptotes for a vertical hyperbola are given by the lines \(y - k = \pm \frac{a}{b}(x - h)\). Substituting \(a = 5\), \(b = 1\), \(h = -3\), and \(k = 1\), the asymptote equations are \(y - 1 = 5(x + 3)\) and \(y - 1 = -5(x + 3)\).

Step 6: Sketch the Hyperbola

To sketch, plot the center \((-3, 1)\), vertices \((-3, 6)\) and \((-3, -4)\), and draw the asymptotes \(y = 5x + 16\) and \(y = -5x - 14\). The hyperbola curves outward from the vertices along the directions outlined by the asymptotes.

Key Concepts

Equation of Hyperbola

The equation of a hyperbola in standard form is crucial for understanding its characteristics. For a vertical hyperbola, like in the exercise, the standard form is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Here, \( (h, k) \) represents the center of the hyperbola. This exercise provides us with the equation \( \frac{(y-1)^{2}}{25}-(x+3)^{2}=1 \). When we compare this with the standard form, we identify \( h = -3 \) and \( k = 1 \), meaning the center of the hyperbola is at \((-3, 1)\).

Understanding the form of the equation helps determine other essential components, such as the vertices, foci, and asymptotes, by using the values of \( a \) and \( b \). Here, \( a^2 = 25 \), so \( a = 5 \), and \( b^2 = 1 \), so \( b = 1 \). Recognizing these components in the equation makes it easier to visualize and graph the hyperbola.

Vertices of Hyperbola

Vertices are the points where the hyperbola intersects its transverse axis. In the case of a vertical hyperbola, the vertices are \( a \) units above and below the center. Referring to the provided hyperbola equation, \( a^2 = 25 \), hence \( a = 5 \).

The center of the hyperbola is at \((-3, 1)\). Therefore, the vertices can be found by adding and subtracting \( a \) from the \( y \)-coordinate of the center.
This calculation results in the vertices at:

• \((-3, 1 + 5) = (-3, 6)\)
• \((-3, 1 - 5) = (-3, -4)\)

These points are crucial in defining the shape and orientation of the hyperbola and help when sketching its graph.

Asymptotes of Hyperbola

Asymptotes are diagonal lines that define the shape and direction the hyperbola will approach but never touch. For a vertical hyperbola, the equation for the asymptotes is given by \( y - k = \pm \frac{a}{b}(x - h) \).

Using the values from the exercise, \( a = 5 \) and \( b = 1 \), with the center at \( (h, k) = (-3, 1) \), we can substitute into the asymptote equation. This yields:

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

Rearranging these gives the simplified asymptote equations:

• \( y = 5x + 16 \)
• \( y = -5x - 14 \)

These lines are crucial to accurately sketching the hyperbola because the hyperbola approaches these lines at infinity.

Foci of Hyperbola

The foci of a hyperbola are important points that help define its shape. They are located \( c \) units from the center, along the transverse axis of the hyperbola. In a vertical hyperbola, the foci lie on the vertical line passing through the center.

The value of \( c \) is calculated using the formula \( c = \sqrt{a^2 + b^2} \). With \( a^2 = 25 \) and \( b^2 = 1 \), we find \( c = \sqrt{26} \). Thus, the foci are located at:

• \((-3, 1 + \sqrt{26})\)
• \((-3, 1 - \sqrt{26})\)

Finding the foci is essential for understanding the full set of properties of a hyperbola and ensuring that its shape and the graph are correctly depicted.