Problem 27 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ \frac{(x+6)^{2}}{36}-\frac{(y+3)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Vertices: (0, -3) and (-12, -3); Foci: (-6 ± 3√5, -3); Asymptotes: y = 1/2x - 6, y = -1/2x - 9.

Step 1: Identify the Standard Form of a Hyperbola Equation

The equation given is \( \frac{(x+6)^{2}}{36} - \frac{(y+3)^{2}}{9} = 1 \), which matches the standard form of a hyperbola centered at \((h, k)\) with a horizontal transverse axis: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Here, \( h = -6 \), \( k = -3 \), \( a^2 = 36 \), and \( b^2 = 9 \). Therefore, \( a = 6 \) and \( b = 3 \).

Step 2: Determine the Coordinates of the Center

The center \((h, k)\) of the hyperbola is \((-6, -3)\) from the equation \( \frac{(x+6)^2}{36} - \frac{(y+3)^2}{9} = 1 \).

Step 3: Find the Vertices

The vertices of the hyperbola are \( a \) units away from the center along the transverse axis. Since the hyperbola opens horizontally, the vertices are at \( (h \pm a, k) \). Thus, the vertices are \((-6+6, -3) = (0, -3)\) and \((-6-6, -3) = (-12, -3)\).

Step 4: Calculate the Foci

The foci are \( c \) units away from the center, where \( c = \sqrt{a^2 + b^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \). Thus, the foci are at \((h \pm c, k) = (-6 \pm 3\sqrt{5}, -3)\). So the coordinates are \((-6 + 3\sqrt{5}, -3)\) and \((-6 - 3\sqrt{5}, -3)\).

Step 5: Write the Equations of the Asymptotes

The asymptotes of a hyperbola with a horizontal transverse axis are given by the equations \( y - k = \pm \frac{b}{a}(x - h) \). Substituting the values, we have \( y + 3 = \pm \frac{3}{6}(x + 6) \). Simplifying, the equations of the asymptotes are \( y + 3 = \pm \frac{1}{2}(x + 6) \), or equivalently, \( y = \frac{1}{2}x - 6 \) and \( y = -\frac{1}{2}x - 9 \).

Step 6: Graph the Hyperbola

To graph the hyperbola, plot the center at \((-6, -3)\), the vertices at \((0, -3)\) and \((-12, -3)\), and the foci at \((-6 + 3\sqrt{5}, -3)\) and \((-6 - 3\sqrt{5}, -3)\). Draw the asymptotes using the equations \( y = \frac{1}{2}x - 6 \) and \( y = -\frac{1}{2}x - 9 \), which intersect at the center. Sketch the hyperbola opening left and right to follow these asymptotes.

Key Concepts

Vertices

In a hyperbola, the vertices are crucial as they define the ends of the hyperbola's transverse axis. Imagine the transverse axis like a bridge that connects the branches of the hyperbola. The vertices fall on this imaginary bridge. They are each located at a distance of "a" units from the center—specifically, along the direction of the transverse axis.
In the given equation \( \frac{(x+6)^{2}}{36}-\frac{(y+3)^{2}}{9}=1 \), the transverse axis is horizontal. So, the vertices can be found by adjusting the x-coordinate of the center, which is \((-6, -3)\). Therefore, the vertices are at \((0, -3)\) and \((-12, -3)\). These points mark the key turning points of the hyperbola.

Asymptotes

Asymptotes are invisible guides that show the path the branches of the hyperbola take as they extend infinitely. They never quite touch the hyperbola but shape its trajectory. For a hyperbola with a horizontal transverse axis like our equation, the asymptotes intersect at the hyperbola’s center and follow this formula:

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

Substituting \(a = 6\), \(b = 3\), \(h = -6\), and \(k = -3\) from the equation, we derive the asymptotes as:

• \( y + 3 = \pm \frac{1}{2}(x + 6) \).

Simplified, they are \( y = \frac{1}{2}x - 6 \) and \( y = -\frac{1}{2}x - 9 \). These lines will guide you when graphing the hyperbola.

Foci

The foci are points of intense interest in a hyperbola because they influence its shape. Unlike the vertices, foci are located outside the actual branches but heavily dictate how wide they are. For any hyperbola, these points lie along the transverse axis and are \(c\) units away from the center, where \(c\) is calculated by \(c = \sqrt{a^2 + b^2}\). For our given hyperbola, \(c\) computes to \(3\sqrt{5}\) or roughly 6.7 when approximated.
The foci therefore land at \((-6 + 3\sqrt{5}, -3)\) and \((-6 - 3\sqrt{5}, -3)\). These points ensure the hyperbola's curves stretch out correctly.

Graphing Hyperbolas

Graphing hyperbolas involves first noting important points: the center, vertices, and foci. Start by marking the center at \((-6, -3)\). Next, plot the vertices at \((0, -3)\) and \((-12, -3)\). These points show where each branch begins its dramatic curve outward. Afterward, plot the approximate foci. Use them to mentally check the hyperbola’s "spread out" nature. The last step includes drawing the asymptotes \( y = \frac{1}{2}x - 6 \) and \( y = -\frac{1}{2}x - 9 \); they create an "X" across the hyperbola. Finally, sketch the branches of the hyperbola to follow these asymptotes without ever touching them. By the end, you'll have a complete hyperbola with all its defining features represented graphically.