Problem 37 - Beginning and Intermediate Algebra

Question

Identify the center of each hyperbola and graph the equation. $$9 x^{2}-y^{2}=36$$

Step-by-Step Solution

Answer

The center of the hyperbola with the given equation \(9x^2 - y^2 = 36\) is at the point (0, 0). To graph the hyperbola, calculate the values of a and b from \(a^2 = 4\) and \(b^2 = 36\), obtaining \(a = 2\) and \(b = 6\). Plot the center, vertices at (-2, 0) and (2, 0), and asymptotes with slopes ±3. Sketch the hyperbola using these points and asymptotes.

Step 1: Write the general equation of a hyperbola

The standard form of the equation of a hyperbola with a horizontal transverse axis and center at the point (h, k) is: \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\]

Step 2: Identify the values of a, b, h, and k in the given equation

We need to rewrite the given equation in the form of the standard equation of a hyperbola. First, divide both sides of the equation by 36 to make the right side equal to 1: \[\frac{9x^2}{36} - \frac{y^2}{36} = 1\] Now rewrite it as follows: \[\frac{x^2}{4} - \frac{(y - 0)^2}{36} = 1\] Comparing it to the standard form, we can identify the values: \(a^2 = 4\), \(b^2 = 36\), \(h = 0\), and \(k = 0\).

Step 3: Calculate the coordinates of the center

The center of the hyperbola is given by the point (h, k). From the previous step, we know that \(h = 0\) and \(k = 0\). Therefore, the center of the hyperbola is at the point (0, 0).

Step 4: Graph the hyperbola

To graph the hyperbola, we need to calculate the values of a and b. We know that \(a^2 = 4\) and \(b^2 = 36\), so: \[a = \sqrt{4} = 2\] \[b = \sqrt{36} = 6\] We also need to calculate the distance between the vertices, which is given by \(2a = 4\). Now, to graph the hyperbola, follow these steps: 1. Mark the center of the hyperbola at point (0, 0) on the coordinate plane. 2. Plot the vertices on the horizontal axis at points (-2, 0) and (2, 0) since a = 2. 3. Plot the points on the vertical axis where the hyperbola approaches the asymptotes at (0, -6) and (0, 6) since b = 6. 4. Draw the asymptotes passing through the center and connecting the vertices. For this hyperbola, they will have slopes of ±3. 5. Sketch the hyperbola using the center, vertices, and asymptotes. The graph of the hyperbola is now complete with the center at the point (0, 0).

Key Concepts

Equation of a Hyperbola

In mathematics, a hyperbola is a type of conic section formed when a plane intersects both nappes of a double cone. The general form of a hyperbola's equation is important as it helps us understand its properties and graph it correctly. The equation of a hyperbola describes its basic shape and orientation on a coordinate plane. It is a vital starting point when analyzing a hyperbola.For hyperbolas, the equation is either - \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\] for a horizontal transverse axis, or- \[\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1\] for a vertical transverse axis. Understanding whether the hyperbola opens horizontally or vertically helps in correctly plotting its graph. Knowing the standard equation allows us to determine other key features like the center, vertices, and asymptotes by comparing our equation to the standard form.

Standard Form

The standard form of a hyperbola is essential for identifying its defining characteristics at a glance. For the equation \[\frac{9x^2}{36} - \frac{y^2}{36} = 1\], we must rearrange it into the standard form \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\]. Doing so simplifies the identification of all necessary elements to graph the hyperbola accurately.- Convert the given equation into this standard form. Divide both sides by 36 resulting in \[\frac{x^2}{4} - \frac{y^2}{36} = 1\].- From here, observe that \(a^2 = 4\), meaning \(a = 2\), and \(b^2 = 36\), meaning \(b = 6\).The standard form enables the easy identification of a hyperbola's orientation and characteristics, which are crucial for graphing or further mathematical analysis.

Center of Hyperbola

The center of a hyperbola is the point from which we calculate symmetry, vertices, and asymptotes. It acts like the anchor point of the hyperbola. By understanding the center, we can clearly plot the hyperbola on a graph.To find the center of a hyperbola, look at the elements \((h, k)\) in the standard form equation. For our current equation, \(h\) and \(k\) both equal 0, showing that the center is at point (0, 0). Identifying the center is critical because the entire hyperbola is symmetric around this point, and it also serves as the base for the asymptotes.

Vertices

Vertices are the nearest points on the hyperbola to the center, and they lie along the transverse axis. They mark the tips of the hyperbola's shape and depict how wide it opens.In the equation \(\frac{x^2}{4} - \frac{y^2}{36} = 1\), since \(a = 2\), the vertices are \(2a\) units apart along the horizontal axis at the points (-2, 0) and (2, 0). By plotting these on the graph, you establish the "ends" of the hyperbola. Vertices are fundamental because they provide a framework for sketching the hyperbola's opening direction and give insight into its extent along the transverse axis.

Asymptotes

Asymptotes of a hyperbola are crucial lines that the curve approaches but never touches. They serve as boundary guides for sketching the hyperbola precisely.For the hyperbola \(\frac{x^2}{4} - \frac{y^2}{36} = 1\), the asymptotes can be found using the slopes \(\pm \frac{b}{a}\). Here, \(b = 6\) and \(a = 2\), which gives slopes \(\pm 3\). Thus, the equations for the asymptotes are \(y = 3x\) and \(y = -3x\). Plot these lines through the center (0, 0).Understanding and drawing asymptotes is vital because they frame the hyperbola's shape and direction, guiding how you graph the curve.