Problem 26 - 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. $$ \frac{(x+1)^{2}}{4}-\frac{(y+3)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Vertices: \((-3, -3)\), \((1, -3)\); Foci: \((-1-\sqrt{13}, -3)\), \((-1+\sqrt{13}, -3)\); Asymptotes: \( y = -3 \pm \frac{3}{2}(x+1) \).

Step 1: Identify the Standard Form

The equation given is \( \frac{(x+1)^2}{4} - \frac{(y+3)^2}{9} = 1 \). This matches the standard form of a hyperbola with a horizontal transverse axis: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), where \((h,k)\) is the center.

Step 2: Identify Center, a, and b

From the equation, the center \((h, k)\) is \((-1, -3)\). Here, \(a^2 = 4\) so \(a = 2\), and \(b^2 = 9\) so \(b = 3\).

Step 3: Find the Vertices

The vertices, for a hyperbola with horizontal transverse axis, are located \(a\) units from the center along the x-axis. So the vertices are \((-1-2, -3) = (-3, -3)\) and \((-1+2, -3) = (1, -3)\).

Step 4: Find the Foci

The foci are found using \(c^2 = a^2 + b^2\). Therefore, \(c^2 = 4 + 9 = 13\). So \(c = \sqrt{13}\). The foci are \((-1 - \sqrt{13}, -3)\) and \((-1 + \sqrt{13}, -3)\).

Step 5: Determine the Asymptotes

The equations of the asymptotes for a hyperbola with horizontal transverse axis are given by \( y = k \pm \frac{b}{a}(x-h) \). Substituting the values, we get the asymptotes: \( y = -3 \pm \frac{3}{2}(x+1) \). These simplify to \( y = -3 + \frac{3}{2}(x+1) \) and \( y = -3 - \frac{3}{2}(x+1) \).

Step 6: Graph the Hyperbola

Plot the center at \((-1, -3)\), the vertices at \((-3, -3)\) and \((1, -3)\), locate the foci at \((-1 - \sqrt{13}, -3)\) and \((-1 + \sqrt{13}, -3)\), and draw the asymptotes \( y = -3 + \frac{3}{2}(x+1) \) and \( y = -3 - \frac{3}{2}(x+1) \). Then sketch the hyperbola opening left and right.

Key Concepts

Vertices of a Hyperbola

In the context of a hyperbola, vertices are pivotal points that represent the closest locations on the hyperbola to its center. Understanding the vertices is essential in sketching the shape of a hyperbola. For the hyperbola given by the equation \( \frac{(x+1)^2}{4} - \frac{(y+3)^2}{9} = 1 \), identifying the vertices is straightforward once you know the equation's structure. Start by finding the center of the hyperbola, which in our case is at \((-1, -3)\). Next, look at \(a\), which is derived from the term \(a^2 = 4\), giving \(a = 2\). Because this hyperbola has a horizontal transverse axis, the vertices are located \(a\) units horizontally away from the center. This means they are at \((-1-a, -3)\) and \((-1+a, -3)\).

• Vertex 1 is at \((-3, -3)\)
• Vertex 2 is at \((1, -3)\)

This positioning creates a clearer picture of where the hyperbola broadens outward along the x-axis.

Foci of a Hyperbola

Foci are another critical component in understanding hyperbolas. They are points situated on the hyperbola's major axis that are farther away from the center than the vertices. To find the foci, use the relationship \(c^2 = a^2 + b^2\). In this exercise, we've calculated \(a^2 = 4\) and \(b^2 = 9\), giving us \(c^2 = 13\). Thus, \(c = \sqrt{13}\). Determining the exact placement of these points helps define the hyperbola's width and the way it "opens". For this horizontal hyperbola:

• Focus 1 is at \((-1 - \sqrt{13}, -3)\)
• Focus 2 is at \((-1 + \sqrt{13}, -3)\)

The foci aren't points you'll directly plot on a basic graph but are crucial for the hyperbola's underlying geometric and analytic properties.

Asymptotes of a Hyperbola

Asymptotes are lines that the hyperbola approaches but never actually intersects. They craft the boundaries within which the hyperbola expands. For a hyperbola like the one in our exercise with a horizontal transverse axis, the asymptotes can be calculated using the formula: \[ y = k \pm \frac{b}{a}(x - h) \]With the given center \((-1, -3)\), and knowing \(a = 2\) and \(b = 3\), the equations for the asymptotes become:

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

The slope \(\frac{b}{a}\) guides the steepness of the asymptote lines. These lines are handy tools in sketching a hyperbola because they define the angle at which the hyperbola's branches open and give visual boundaries that frame the graph, showing clearly where the curves will never cross.