Problem 33 - Algebra and Trigonometry

Question

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 (-4, -3), the vertices are (-1, -3) and (-7, -3), the asymptotes are \(y = -3 ± \frac{4}{3}(x+4)\), and the foci are (1,-3) and (-9,-3).

Step 1: Identify the center

The center of the hyperbola is given as the inverse of the values \('h'\) and \('k'\) in the equation \(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\). Here, \(h = -4\) and \(k = -3\). Hence, the center is \((-4, -3)\).

Step 2: Find the vertices and asymptotes

The vertices of the hyperbola are calculated by adding and subtracting \('a'\) from the \('x'\)-coordinate of the center. The asymptotes of the hyperbola are given by the equations \(k ± \frac{b}{a}(x-h) = y\). Here, \(a = \sqrt{9} = 3\) and \(b = \sqrt{16} = 4\). Hence, the vertices are \((-4+3, -3) = (-1, -3)\) and \((-4-3, -3) = (-7, -3)\), and the asymptotes are \(y = -3 ± \frac{4}{3}(x+4)\).

Step 3: Calculate the foci

The foci of the hyperbola are determined by adding and subtracting \('c'\) from the \('x'\)-coordinate of the center, where \(c=\sqrt{a^{2}+b^{2}}\). Here, \(a = 3\) and \(b = 4\). Hence, \(c = \sqrt{3^{2}+4^{2}} = \sqrt{25} = 5\), and the foci are \((-4+5, -3) = (1, -3)\) and \((-4-5, -3) = (-9, -3)\).

Key Concepts

Equations of Asymptotes

In a hyperbola, the asymptotes are crucial as they define the slanting lines that intersect the center of the hyperbola and guide its shape. Asymptotes are particularly important because, as the curves of a hyperbola extend, they get infinitely close to these lines without ever touching them. The equation of a hyperbola in standard form is given by \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Here, \(h\) and \(k\) are the coordinates of the center, while \(a\) and \(b\) are the distances along the \(x\) and \(y\) axes, respectively. The equations of the asymptotes for a hyperbola centered at \((h, k)\) are given by the following formulas, reflecting the slopes of the asymptotes:

• \( y - k = \pm \frac{b}{a} (x - h) \) for horizontal hyperbolas.
• \( y - k = \pm \frac{a}{b} (x - h) \) for vertical hyperbolas.

For the given exercise, \( h = -4 \), \( k = -3 \), \( a = 3 \), and \( b = 4 \), the hyperbola is horizontal. Thus, substituting these values into the asymptote equation \( y = k \pm \frac{b}{a}(x-h) \), we find:

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

These equations reveal the asymptotes' paths and provide a framework for graphing the hyperbola.

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. This axis is the main "line" along which the hyperbola opens. For hyperbolas in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the transverse axis is horizontal.The vertices are determined by the distance \( a \), which is the same as the semi-major axis of the ellipse counterpart. They are found by calculating:

• \((h \pm a, k)\)

In the exercise, for a center at \((-4, -3)\) and \(a = 3\), adding and subtracting \(a\) from the \(x\)-coordinate gives us the vertices:

• \((-4 + 3, -3) = (-1, -3)\)
• \((-4 - 3, -3) = (-7, -3)\)

These points are critical because they give the hyperbola its initial shape and orientation on the graph.

Foci of Hyperbola

The foci of a hyperbola are points located along the transverse axis, which have a unique property related to the hyperbola's shape. Specifically, the difference in distances from any point on the hyperbola to each focus is a constant.To find the foci of a hyperbola given by the equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), you need to use the formula:

• \( c = \sqrt{a^2 + b^2} \)

The foci lie \( c \) units away from the center, so their coordinates are \((h \pm c, k)\).For the exercise, we calculate:

• \( a = 3, b = 4 \)
• \( c = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \)

Therefore, the foci are found at:

• \((-4 + 5, -3) = (1, -3)\)
• \((-4 - 5, -3) = (-9, -3)\)

These focus points help us understand the hyperbola's stretched shape and are essential in applications like GPS and astronomy.