Hyperbola Analysis Finding Center, Eccentricity, Vertices, Foci, And Axes
The intriguing world of conic sections introduces us to the hyperbola, a captivating curve defined by its unique properties. Delving into the hyperbola's anatomy involves pinpointing key features like its center, eccentricity, vertices, foci, directrix, and the lengths of its axes. This article serves as a comprehensive guide, meticulously dissecting the equation $9x^2 - 16y^2 - 18x - 64y - 199 = 0$ to unveil these fundamental characteristics.
Transforming the Equation into Standard Form
To decipher the hyperbola's secrets, we must first transform the given equation into its standard form. This involves a process called completing the square. Our initial equation is $9x^2 - 16y^2 - 18x - 64y - 199 = 0$. Let's begin by grouping the x and y terms together: $(9x^2 - 18x) + (-16y^2 - 64y) = 199$. Next, factor out the coefficients of the squared terms: $9(x^2 - 2x) - 16(y^2 + 4y) = 199$. Now, we complete the square for both the x and y expressions. For the x terms, we take half of the coefficient of the x term (-2), square it (1), and add it inside the parenthesis. We must also add 9 times this value to the right side to maintain the equation's balance. For the y terms, we take half of the coefficient of the y term (4), square it (4), and add it inside the parenthesis. Similarly, we subtract 16 times this value from the right side. This yields: $9(x^2 - 2x + 1) - 16(y^2 + 4y + 4) = 199 + 9 - 64$. Now we can rewrite the expressions in parentheses as squared terms: $9(x - 1)^2 - 16(y + 2)^2 = 144$. Finally, we divide both sides by 144 to obtain the standard form of the hyperbola's equation: $\frac{(x - 1)^2}{16} - \frac{(y + 2)^2}{9} = 1$. This form provides valuable insights into the hyperbola's key attributes, setting the stage for our exploration.
Identifying the Hyperbola's Center
The center of a hyperbola, a pivotal point in its geometry, serves as the midpoint of both the transverse and conjugate axes. In the standard form equation, $\frac{(x - h)2}{a2} - \frac{(y - k)2}{b2} = 1$, the center is represented by the coordinates (h, k). Comparing this to our transformed equation, $\frac{(x - 1)^2}{16} - \frac{(y + 2)^2}{9} = 1$, we can readily identify the center as (1, -2). This point acts as the hyperbola's central anchor, around which all other features are defined. Locating the center is a crucial first step in understanding the hyperbola's spatial orientation and overall structure. It provides a reference point for determining the positions of vertices, foci, and the directrix, as well as the lengths and orientations of the axes. Accurately determining the center is therefore paramount for a complete analysis of the hyperbola.
Calculating the Eccentricity
Eccentricity, a defining characteristic of conic sections, quantifies the hyperbola's deviation from a circular shape. For hyperbolas, eccentricity (e) is always greater than 1. It is calculated using the formula $e = \fracc}{a}$, where c represents the distance from the center to each focus, and a represents the distance from the center to each vertex. To find c, we use the relationship $c^2 = a^2 + b^2$, where a and b are the lengths of the semi-major and semi-minor axes, respectively. From our standard equation, $\frac{(x - 1)^2}{16} - \frac{(y + 2)^2}{9} = 1$, we can identify $a^2 = 16$ and $b^2 = 9$. Therefore, a = 4 and b = 3. Plugging these values into the equation for c, we get $c^2 = 16 + 9 = 25$, so c = 5. Now, we can calculate the eccentricity{4}$. This value, greater than 1, confirms that we are indeed dealing with a hyperbola. The eccentricity provides valuable information about the hyperbola's shape; a higher eccentricity indicates a more elongated hyperbola, while a value closer to 1 suggests a less pronounced curve. Understanding the eccentricity is crucial for visualizing the hyperbola's overall form and its relationship to other conic sections.
Pinpointing the Vertices
The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis, the axis that passes through the foci and the center. In our standard equation, $\frac{(x - 1)^2}{16} - \frac{(y + 2)^2}{9} = 1$, the transverse axis is horizontal because the term with the positive sign is associated with the x variable. The distance from the center to each vertex is given by a, which we found to be 4. Since the center is at (1, -2), the vertices are located 4 units to the left and right of the center along the horizontal axis. Therefore, the vertices are at (1 + 4, -2) = (5, -2) and (1 - 4, -2) = (-3, -2). The vertices are key points in defining the hyperbola's shape and orientation. They mark the closest points of the hyperbola to its center and serve as the endpoints of the transverse axis. Their accurate determination is essential for sketching the hyperbola and understanding its geometric properties.
Locating the Foci
The foci (plural of focus) are two fixed points within the hyperbola that play a crucial role in its definition. The distance from the center to each focus is denoted by c, which we calculated to be 5. Since the transverse axis is horizontal, the foci are located 5 units to the left and right of the center along this axis. With the center at (1, -2), the foci are at (1 + 5, -2) = (6, -2) and (1 - 5, -2) = (-4, -2). The foci are fundamental to the hyperbola's geometric definition: for any point on the hyperbola, the absolute difference in distances to the two foci is constant. This property distinguishes the hyperbola from other conic sections. The position of the foci significantly influences the hyperbola's shape; the farther the foci are from the center, the more elongated the hyperbola becomes.
Determining the Equation of the Directrix
The directrix is a line associated with the hyperbola, and its relationship to the foci defines the curve. For a hyperbola, there are two directrices, each located at a distance of a/e from the center, on either side. In our case, a = 4 and e = 5/4, so the distance from the center to each directrix is $4 / (5/4) = 4 * (4/5) = 16/5$. Since the transverse axis is horizontal, the directrices are vertical lines. The equations of the directrices are given by $x = h ± \frac{a}{e}$, where (h, k) is the center. Plugging in the values, we get $x = 1 ± \frac{16}{5}$. Thus, the equations of the directrices are $x = 1 + \frac{16}{5} = \frac{21}{5}$ and $x = 1 - \frac{16}{5} = -\frac{11}{5}$. The directrices, along with the foci, provide an alternative definition of the hyperbola: for any point on the hyperbola, the ratio of its distance to a focus to its distance to the corresponding directrix is equal to the eccentricity. This property highlights the interconnectedness of these elements in shaping the hyperbola.
Calculating the Lengths of the Axes
The hyperbola possesses two axes: the transverse axis and the conjugate axis. The transverse axis is the axis that passes through the foci and vertices, while the conjugate axis is perpendicular to the transverse axis and passes through the center. The length of the transverse axis is 2a, and the length of the conjugate axis is 2b. In our case, a = 4 and b = 3. Therefore, the length of the transverse axis is 2 * 4 = 8 units, and the length of the conjugate axis is 2 * 3 = 6 units. The lengths of the axes provide a measure of the hyperbola's dimensions. The transverse axis dictates the hyperbola's spread along its major direction, while the conjugate axis influences its width. Understanding these lengths is crucial for accurately sketching and visualizing the hyperbola.
Conclusion
By meticulously analyzing the equation $9x^2 - 16y^2 - 18x - 64y - 199 = 0$, we have successfully unveiled the hyperbola's key characteristics. We determined the center to be (1, -2), the eccentricity to be 5/4, the vertices to be (5, -2) and (-3, -2), the foci to be (6, -2) and (-4, -2), the equations of the directrices to be $x = \frac{21}{5}$ and $x = -\frac{11}{5}$, and the lengths of the transverse and conjugate axes to be 8 and 6 units, respectively. This comprehensive analysis not only illuminates the specific properties of this hyperbola but also provides a framework for understanding the broader characteristics of these fascinating conic sections. Mastering the techniques of identifying and calculating these parameters is essential for anyone seeking a deeper understanding of conic sections and their applications in various fields, from mathematics and physics to engineering and architecture.
