Problem 29 - Beginning and Intermediate Algebra

Question

Identify the center of each hyperbola and graph the equation. $$\frac{(x-2)^{2}}{9}-\frac{(y+3)^{2}}{16}=1$$

Step-by-Step Solution

Answer

The center of the given hyperbola is (2, -3). The vertices are (5, -3) and (-1, -3), and the foci are (7, -3) and (-3, -3). The asymptotes are given by the equations: \(y = -3 + \frac{4}{3}(x - 2)\) and \(y = -3 - \frac{4}{3}(x - 2)\). To graph the hyperbola, plot the center, vertices, foci, and draw the asymptotes. Then, sketch the hyperbola by drawing curves that approach the asymptotes and pass through the vertices.

Step 1: Identify the standard form of a hyperbola equation

A hyperbola equation is given by the following standard form: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] where (h,k) is the center of the hyperbola.

Step 2: Find the center of the given hyperbola

For the given hyperbola equation: \[\frac{(x-2)^2}{9}-\frac{(y+3)^2}{16}=1\] We can see that it is in the standard form, with h = 2, k = -3. Therefore, the center of the hyperbola is (2, -3).

Step 3: Calculate the vertices, foci, and asymptotes

Comparing the given equation to the standard form, we can identify that a^2 = 9 and b^2 = 16. Therefore, we can find a and b values as follows: a = √9 = 3 b = √16 = 4 The vertices are located at (h±a, k), so for our hyperbola: Vertices: (2±3, -3) => (5, -3) and (-1, -3) To find the foci, we need to calculate the distance from the center using c, defined as c = √(a^2 + b^2): c = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 The foci are located at (h±c, k): Foci: (2±5, -3) => (7, -3) and (-3, -3) To find the equation of the asymptotes, we use the following expressions: \(y = k \pm \frac{b}{a}(x - h)\) For our hyperbola: Asymptote 1: \(y = -3 + \frac{4}{3}(x - 2)\) Asymptote 2: \(y = -3 - \frac{4}{3}(x - 2)\)

Step 4: Graph the hyperbola, vertices, foci, and asymptotes

Now that we have all the necessary components, we can graph the hyperbola. 1. Plot the center: (2, -3) 2. Graph the vertices: (5, -3) and (-1, -3) 3. Graph the foci: (7, -3) and (-3, -3) 4. Draw the asymptotes using their equations. 5. Sketch the hyperbola by drawing a curve along the asymptotically approaching asymptotes, passing through the vertices. Our final graph should show the hyperbola, center, vertices, foci, and asymptotes.

Key Concepts

Center of Hyperbola

The center of a hyperbola is the point that serves as the midpoint of the line segment joining its vertices. It is crucial because it acts as a reference point for positioning the entire hyperbola on the coordinate plane. In the standard form of a hyperbola's equation \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\], \((h, k)\) represents the center.

• In our example, the equation is \(\frac{(x-2)^2}{9} - \frac{(y+3)^2}{16} = 1\).
• By comparing this with the standard form, we identify \(h = 2\) and \(k = -3\).
• Thus, the center of the hyperbola is at \((2, -3)\).

Understanding the center helps in determining the placement and symmetry of a hyperbola on the graph. It is like finding the 'origin' around which the hyperbola is structured.

Vertices of Hyperbola

The vertices of a hyperbola are located at the points where the hyperbola intersects its transverse axis, which is the line passing through the center and defined by the equation. They signify the closest points of the hyperbola to the center, unlike the foci which are located further away.

• To find the vertices, use the formula \((h \pm a, k)\) for a horizontal hyperbola.
• In our specific equation with \(a = 3\), \(h = 2\), and \(k = -3\), the vertices are calculated as \((2\pm3, -3)\).
• This results in the vertices being \((5, -3)\) and \((-1, -3)\).

These points are essential when sketching the shape of the hyperbola, as they mark the preliminary guiding frameworks. The transverse axis, marked by these vertices, sets the stage for graphing the proper curve.

Foci of Hyperbola

The foci of a hyperbola are internal points located on the transverse axis, beyond the vertices. These points are fundamental because they determine the shape and orientation of the hyperbola, by being the loci towards which the arms of the hyperbola tend. The distance from the center to each focus is given by \(c\), where \(c = \sqrt{a^2 + b^2}\).

• In our scenario, \(a = 3\) and \(b = 4\), thus \(c = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\).
• The foci are then positioned at \((h \pm c, k)\).
• For our hyperbola: \((2\pm5, -3)\), which locates the foci at \((7, -3)\) and \((-3, -3)\).

The distance between each arm of the hyperbola and its respective focus remains approximately constant, creating the characteristic hyperbolic shape.

Equation of Asymptotes

Asymptotes of a hyperbola are lines that the hyperbolic curves approach but never actually meet. They define the bounds for the hyperbola's curves, offering a boundary guide while sketching. The equation for the asymptotes of a hyperbola with a horizontal transverse axis is \[y = k \pm \frac{b}{a}(x - h)\].

• For the hyperbola in our problem, we have \(b = 4\), \(a = 3\), \(h = 2\), and \(k = -3\).
• Thus the equations become: \(y = -3 + \frac{4}{3}(x - 2)\) and \(y = -3 - \frac{4}{3}(x - 2)\).

These equations guide how the hyperbola opens and stretches, serving as critical references for drawing the exact paths of the hyperbola on a graph. Understanding asymptotes is key to accurately representing the stretched shape of the hyperbola.