Problem 12 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ \frac{\left(y-\frac{1}{4}\right)^{2}}{4}-\frac{(x+3)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Center: \((-3, \frac{1}{4})\), Vertices: \((-3, \frac{9}{4})\), \((-3, -\frac{7}{4})\), Asymptotes: \(y = \frac{2}{3}x + 2\), \(y = -\frac{2}{3}x - 1.5\). Eccentricity: \(\frac{\sqrt{13}}{2}\).

Step 1: Identify the Standard Form of the Hyperbola

The given equation \(\frac{(y-\frac{1}{4})^{2}}{4} - \frac{(x+3)^{2}}{9} = 1\) is in the standard form for a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). From this, we identify \( h = -3 \), \( k = \frac{1}{4} \), \( a^2 = 4 \), and \( b^2 = 9 \).

Step 2: Find the Center of the Hyperbola

The center \((h, k)\) of the hyperbola is \((-3, \frac{1}{4})\).

Step 3: Determine the Vertices

For a vertical hyperbola, vertices are located \(a\) units up and down from the center along the \(y\)-axis. Here, \(a = \sqrt{4} = 2\). Thus, the vertices are at \((-3, \frac{1}{4} + 2)\) and \((-3, \frac{1}{4} - 2)\), which simplifies to \((-3, \frac{9}{4})\) and \((-3, -\frac{7}{4})\).

Step 4: Determine the Foci

The distance of foci from the center \(c\) is found using \(c^2 = a^2 + b^2\). Here, \(c^2 = 4 + 9 = 13\) so \(c = \sqrt{13}\). The foci are at \((-3, \frac{1}{4} + \sqrt{13})\) and \((-3, \frac{1}{4} - \sqrt{13})\).

Step 5: Determine the Asymptotes

In this vertical hyperbola, the asymptote equations are \(y = k \pm \frac{a}{b}(x-h)\). So, \(k = \frac{1}{4}\), \(h = -3\), \(a = 2\), \(b = 3\). Therefore, the asymptotes are \(y = \frac{1}{4} \pm \frac{2}{3}(x + 3)\), which simplifies to \(y = \frac{2}{3}x + 2\) and \(y = -\frac{2}{3}x - 1.5\).

Step 6: Calculate the Eccentricity

The eccentricity \(e\) is calculated by \(e = \frac{c}{a}\). Substituting the values, \(e = \frac{\sqrt{13}}{2}\).

Step 7: Graph the Hyperbola

The graph of the hyperbola should be sketched showing the center \((-3, \frac{1}{4})\), vertices \((-3, \frac{9}{4})\) and \((-3, -\frac{7}{4})\), foci, and the asymptotes \(y = \frac{2}{3}x + 2\) and \(y = -\frac{2}{3}x - 1.5\).

Key Concepts

Center of Hyperbola

The center of a hyperbola provides a crucial reference point for determining other key features of the curve. In the standard form equation of a hyperbola, \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1, \]where the hyperbola opens vertically, the center is denoted as \((h, k)\). For the given equation, \[ \frac{(y-\frac{1}{4})^{2}}{4} - \frac{(x+3)^{2}}{9} = 1, \]we identify \( h = -3 \) and \( k = \frac{1}{4} \). Therefore, the center of the hyperbola is located at the point \((-3, \frac{1}{4})\).
This point is critical because it serves as a symmetric anchor point for locating vertices, foci, and developing the asymptotic lines.

Vertices of Hyperbola

In a hyperbola, vertices are the points where the curve intersects its axis of symmetry. For a vertical hyperbola, they lie along the \(y\)-axis at a distance of \(a\) from the center, where \(a\) is the semi-major axis length.
From the equation, we know \(a^2 = 4\), so \(a = \sqrt{4} = 2\).
The vertices are found by adjusting the \(y\)-coordinate of the center by \(a\):

• \((-3, \frac{1}{4} + 2) = (-3, \frac{9}{4})\)
• \((-3, \frac{1}{4} - 2) = (-3, -\frac{7}{4})\)

These vertices provide the most natural and visible markers of the hyperbola's extend in the vertical direction and help with sketching the overall shape.

Asymptotes of Hyperbola

Asymptotes are essential features of hyperbolas that represent the lines that the curve approaches infinitely but never intersects. They provide a skeletal framework showing the hyperbola's overall orientation and direction. For a vertical hyperbola, the asymptote equations take the form:\[ y = k \pm \frac{a}{b}(x-h). \]Here, substituting \(k = \frac{1}{4}\), \(a=2\), \(b=3\), and \(h=-3\):
We derive the equations:

• \(y = \frac{1}{4} + \frac{2}{3}(x + 3) = \frac{2}{3}x + 2\)
• \(y = \frac{1}{4} - \frac{2}{3}(x + 3) = -\frac{2}{3}x - 1.5\)

These asymptotes are lines that help to define the boundary towards which the arms of the hyperbola curve forever extend.

Foci of Hyperbola

The foci of a hyperbola are two special points that are a fixed distance, known as the focal distance, from the center of the hyperbola along its axis of symmetry. The hyperbola bends around these foci.
To find the foci, we use the equation:\[ c^2 = a^2 + b^2, \]where \(c\) is the distance from the center to each focus.
From our values, \(a^2 = 4\) and \(b^2 = 9\), hence \(c^2 = 13\) and \(c = \sqrt{13}\).

• The foci are located at \((-3, \frac{1}{4} + \sqrt{13})\)
• \((-3, \frac{1}{4} - \sqrt{13})\)

These foci are critical for laying out the hyperbola, as each arm bends around and diverges in proximity to one of these points.

Eccentricity of Hyperbola

Eccentricity is a numerical measure that captures how "stretched" a conic section is. For a hyperbola, this value is always greater than 1. It is given by the ratio \[ e = \frac{c}{a}, \]where \(c\) is the distance from the center to a focus, and \(a\) is the distance from the center to a vertex.
From previous calculations, \(c = \sqrt{13}\) and \(a = 2\), so the eccentricity is:\[ e = \frac{\sqrt{13}}{2}. \]The higher this value, the more elongated the hyperbola appears, confirming that it's the most open of all conic sections.