Vertices Of Hyperbola \(\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1\) Are (-1,6) And (-1,-2)

Introduction to Hyperbolas

In the realm of conic sections, hyperbolas stand out as fascinating curves defined by their unique geometric properties. Unlike ellipses, which are formed by slicing a cone at an angle that doesn't intersect the base, hyperbolas arise from a slice that intersects both halves of a double cone. This distinctive intersection leads to the hyperbola's characteristic two-branch structure, each curving away from the other. To truly understand the intricacies of hyperbolas, we must delve into their equations, key parameters, and how these elements dictate the hyperbola's shape and position within the coordinate plane. Central to this understanding is the concept of vertices, which are the points where the hyperbola most closely approaches its center, marking the endpoints of the transverse axis. In this article, we will focus on a specific hyperbola defined by the equation ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$. Our primary goal is to rigorously determine whether the vertices of this hyperbola are indeed at the points ${(-1,6)}$ and ${(-1,-2)}$. By carefully analyzing the equation and applying the standard form of a hyperbola, we will either confirm or refute this claim, providing a comprehensive understanding of how to identify the vertices of a hyperbola. This exploration will not only enhance our grasp of hyperbolas but also reinforce our skills in manipulating and interpreting conic section equations. Understanding the vertices of a hyperbola is crucial for sketching its graph and solving various problems related to its geometry. The vertices, along with other key features such as the center, foci, and asymptotes, provide a complete picture of the hyperbola's shape and orientation. In the following sections, we will break down the equation, identify the key parameters, and then use these parameters to calculate the vertices. This step-by-step approach will ensure a clear and thorough understanding of the process.

Decoding the Hyperbola Equation: ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$

The given equation, ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$, is a quintessential representation of a hyperbola in its standard form. To fully grasp the significance of this equation, we must meticulously dissect its components and understand how each parameter contributes to the hyperbola's overall structure. The standard form of a hyperbola equation provides a wealth of information at a glance, allowing us to determine the hyperbola's orientation, center, and the lengths of its axes. This form is particularly useful because it directly relates the equation's coefficients to the geometric properties of the hyperbola, such as the distance between the vertices and the location of the foci. In this case, the equation immediately reveals that the hyperbola is vertically oriented because the term involving ${y}$ is the positive one. The equation clearly shows the form ${\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1}$, where ${(h, k)}$ represents the center of the hyperbola. By carefully examining our equation, we can identify the values of ${h}$ and ${k}$ as ${-1}$ and ${2}$ respectively. Thus, the center of this hyperbola is at the point ${(-1, 2)}$. The values under the squared terms, ${16}$ and ${144}$, are the squares of the semi-major axis ${a}$ and the semi-minor axis ${b}$ respectively. Since ${16}$ is under the ${(y-2)^2}$ term, we have ${a^2 = 16}$, which implies ${a = 4}$. This value represents the distance from the center to each vertex along the vertical axis. Similarly, ${144}$ is under the ${(x+1)^2}$ term, giving us ${b^2 = 144}$, and thus ${b = 12}$. This value is crucial for determining the shape and asymptotes of the hyperbola. The relationship between ${a}$ and ${b}$ also helps in finding the foci of the hyperbola, which are essential points that define its curvature. By understanding these fundamental parameters, we lay the groundwork for accurately determining the vertices and other critical features of the hyperbola. The process of extracting these values from the equation is a vital skill in the study of conic sections, allowing us to quickly interpret and analyze hyperbolic equations.

Pinpointing the Vertices: A Step-by-Step Calculation

Now that we have successfully decoded the hyperbola equation ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$ and identified its key parameters, we can proceed with the crucial step of determining the vertices. The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis, which is the axis that passes through the center and the foci. For a vertically oriented hyperbola, like the one we are examining, the vertices lie directly above and below the center. This orientation is evident from the positive term in the equation being associated with the ${y}$ variable. To find the vertices, we need to understand their relationship with the center and the semi-major axis length, denoted by ${a}$. As established earlier, the center of our hyperbola is at ${(-1, 2)}$, and the value of ${a}$ is ${4}$. Since the hyperbola is vertically oriented, the vertices will be located ${a}$ units above and below the center along the vertical axis. This means we need to add and subtract ${a}$ from the ${y}$-coordinate of the center while keeping the ${x}$-coordinate constant. Starting with the center ${(-1, 2)}$, we add ${a = 4}$ to the ${y}$-coordinate to find one vertex: ${(-1, 2 + 4) = (-1, 6)}$. This point represents the upper vertex of the hyperbola. Next, we subtract ${a = 4}$ from the ${y}$-coordinate of the center to find the other vertex: ${(-1, 2 - 4) = (-1, -2)}$. This point represents the lower vertex of the hyperbola. Thus, through a clear and methodical calculation, we have determined that the vertices of the hyperbola ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$ are indeed at the points ${(-1, 6)}$ and ${(-1, -2)}$. This confirms the initial statement and provides a solid foundation for further analysis of the hyperbola's properties.

Verifying the Vertices: Graphical and Analytical Confirmation

To ensure the accuracy of our calculations, it is prudent to verify the vertices of the hyperbola both graphically and analytically. Graphical verification involves plotting the hyperbola and visually confirming that the calculated vertices align with the curve's endpoints along the transverse axis. Analytical verification, on the other hand, involves substituting the coordinates of the vertices back into the original equation to confirm that they satisfy the equation. For graphical verification, one can use graphing software or online tools to plot the hyperbola ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$. By plotting the hyperbola, we can visually inspect the curve and identify the points where it intersects the vertical line passing through the center ${x = -1}$. The endpoints of the hyperbola should visually align with the points ${(-1, 6)}$ and ${(-1, -2)}$, confirming our calculated vertices. This graphical approach provides an intuitive and visual confirmation of our results, making it an essential step in the verification process. Now, let's proceed with the analytical verification. To verify that ${(-1, 6)}$ and ${(-1, -2)}$ are indeed the vertices, we substitute these coordinates into the equation ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$. For the point ${(-1, 6)}$, we have: ${\frac{(6-2)^2}{16}-\frac{(-1+1)^2}{144} = \frac{4^2}{16}-\frac{0^2}{144} = \frac{16}{16} - 0 = 1}$. The equation holds true for ${(-1, 6)}$. Next, for the point ${(-1, -2)}$, we have: ${\frac{(-2-2)^2}{16}-\frac{(-1+1)^2}{144} = \frac{(-4)^2}{16}-\frac{0^2}{144} = \frac{16}{16} - 0 = 1}$. The equation also holds true for ${(-1, -2)}$. Both graphical and analytical verifications confirm that our calculated vertices, ${(-1, 6)}$ and ${(-1, -2)}$, are indeed the correct vertices for the given hyperbola. This comprehensive verification process underscores the importance of not only calculating but also validating the results to ensure accuracy and a deep understanding of the hyperbola's properties.

Conclusion: Affirming the Vertices of the Hyperbola

In summary, through a meticulous and step-by-step analysis, we have successfully determined and verified the vertices of the hyperbola defined by the equation ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$. Our journey began with a thorough examination of the equation, where we identified the center of the hyperbola as ${(-1, 2)}$ and the semi-major axis length as ${a = 4}$. This foundational understanding allowed us to calculate the vertices by considering the hyperbola's vertical orientation and the distance ${a}$ from the center along the transverse axis. Our calculations led us to the vertices ${(-1, 6)}$ and ${(-1, -2)}$. To ensure the accuracy of our results, we employed a dual verification approach. First, we used a graphical method, plotting the hyperbola and visually confirming that the calculated vertices aligned perfectly with the curve's endpoints. This visual confirmation provided an intuitive understanding of the hyperbola's shape and the position of its vertices. Second, we performed an analytical verification by substituting the coordinates of the vertices back into the original equation. This rigorous check confirmed that both points satisfy the equation, solidifying our confidence in the calculated vertices. Thus, we can confidently affirm that the vertices for the hyperbola ${\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1}$ are indeed ${(-1, 6)}$ and ${(-1, -2)}$. This exercise not only reinforces our understanding of hyperbolas but also highlights the importance of a methodical approach in solving mathematical problems. By combining analytical calculations with graphical and analytical verification techniques, we can achieve a comprehensive and accurate understanding of conic sections and their properties. This knowledge is invaluable in various fields, including physics, engineering, and computer graphics, where hyperbolas play a significant role in describing various phenomena and shapes. Through this exploration, we have deepened our understanding of hyperbolas and honed our problem-solving skills, paving the way for further explorations in the fascinating world of conic sections.

Answer: A. True