Problem 33 - College Algebra

Question

In Exercises 33-42, use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(x+4)^{2}}{9}-\frac{(y+3)^{2}}{16}=1$$

Step-by-Step Solution

Answer

The center of the hyperbola is at (4,3). The vertices are at (1,3) and (7,3). The equation of the asymptotes is \(y=\pm\frac{4}{3}x+\frac{1}{3}\) and \(y=\pm\frac{4}{3}x+11\). The foci of the hyperbola are at (-1,3) and (9,3).

Step 1: Identify the center

The center of the hyperbola is at \((-h, -k)\) where \(h\) and \(k\) are from the given equation: \(\frac{(x+h)^{2}}{a^{2}}-\frac{(y+k)^{2}}{b^{2}}=1\). Therefore in this case, the center is \((-(-4), -(-3)) = (4, 3)\)

Step 2: Identify the Vertices

The vertices of this horizontal hyperbola, as per its standard form, are \((h \pm a, k)\). Since we recognize this as a horizontal hyperbola (due to positive 'x' term), we identify \(a=\sqrt{9}=3\) with \(a\) being the denominator of the \(x\) term in the equation. Thus, the vertices will be: \((4 \pm 3, 3)\) = \((7, 3)\) and \((1, 3)\)

Step 3: Identify the Asymptotes

The equations for the asymptotes for horizontal hyperbolas are \(y=k\pm\frac{b}{a}(x-h)\), so taking \(b=\sqrt{16}=4\), we get \(y=3\pm\frac{4}{3}(x-4)\) which simplifies to \(y=\pm\frac{4}{3}x+\frac{1}{3}\) and \(y=\pm\frac{4}{3}x+11\)

Step 4: Identify the Foci

The foci lie on the x-axis for horizontal hyperbolas. The distance from the center to either focus (focal distance) is \(\sqrt{a^2+b^2}\). Substituting \(a=3\) and \(b=4\) we have: \(\sqrt{9+16} = \sqrt{25} = 5\). Thus the foci are at \((4 \pm 5, 3)\) = \((-1, 3)\) and \((9, 3)\)

Key Concepts

Center of Hyperbola

Understanding the center of a hyperbola is crucial because it serves as the reference point for locating other key components like vertices and foci. In the standard form of a hyperbola \[ \frac{(x + h)^{2}}{a^{2}} - \frac{(y + k)^{2}}{b^{2}} = 1 \]the center is given by \((-h, -k)\). In this particular equation, \[ \frac{(x + 4)^{2}}{9} - \frac{(y + 3)^{2}}{16} = 1 \]the terms \(h\) and \(k\) are identified as \(-4\) and \(-3\) respectively, hence the center of our hyperbola is located at \((4, 3)\). The center marks the midpoint between the vertices of the hyperbola. It is symmetrical with respect to the hyperbola’s geometry, playing a key role in differentiating between the horizontal and vertical types of hyperbolas. Here, the positive term under the x-component indicates a horizontal hyperbola, confirming the center’s pivotal role.

Vertices of Hyperbola

The vertices of a hyperbola are the points on the graph that are closest to its center. For horizontal hyperbolas, these are found along the x-axis, a distance \(a\) from the center point. In this hyperbola \[ \frac{(x + 4)^{2}}{9} - \frac{(y + 3)^{2}}{16} = 1 \]we find \(a = \sqrt{9} = 3\). Thus, the vertices are determined as \((h \pm a, k)\).

• By calculating \(4 \pm 3\), we find the vertices at \((7, 3)\) and \((1, 3)\).

These vertices provide a visual line joining the primary axis of the hyperbola. They mark the narrowest part of the hyperbola and are equidistant from the center, ensuring the hyperbola opens correctly in the horizontal direction.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are the invisible lines that the hyperbola approaches but never touches. They define the hyperbola’s shape, ensuring the branches open symmetrically relative to these lines. For a horizontal hyperbola in standard form \[ \frac{(x + h)^{2}}{a^{2}} - \frac{(y + k)^{2}}{b^{2}} = 1 \], the asymptotes are given by

• \(y = k \pm \frac{b}{a}(x - h)\)

For our equation, where \(b = \sqrt{16} = 4\) and \(a = 3\), we calculate the asymptotes as:

• \(y = 3 \pm \frac{4}{3}(x - 4)\)

Simplifying these leads to equations:

• \(y = \frac{4}{3}x + \frac{1}{3}\)
• \(y = -\frac{4}{3}x + 11\)

These lines run through the center and direct the opening of the hyperbola wings along their slopes, fulfilling the geometric structure and balance.

Foci of Hyperbola

The foci of a hyperbola are two fixed points situated inside each curve of the hyperbola. They are essential because the difference in distances from any point on the hyperbola to the foci is constant. For a hyperbola given by \[ \frac{(x + h)^{2}}{a^{2}} - \frac{(y + k)^{2}}{b^{2}} = 1 \], the foci are placed at a distance \(\sqrt{a^2 + b^2}\) centered along the transverse axis.For this particular equation, where \(a = 3\) and \(b = 4\), the focus distance is:

• \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Using the center \((4, 3)\), the foci are located at:

• \((4 \pm 5, 3)\), or \((-1, 3)\) and \((9, 3)\)

These foci lie along the path of the x-coordinate in a horizontal hyperbola and serve as the mathematical heart, shaping the curve and enforcing the property of symmetry and constant distance.