Hyperbola Equation Find The Center And Vertices

In the realm of conic sections, the hyperbola stands out with its distinctive two-branch shape, opening outwards in opposite directions. Understanding the properties of a hyperbola, such as its center and vertices, is crucial for accurately graphing and analyzing its behavior. In this comprehensive guide, we will embark on a step-by-step journey to dissect the equation of a hyperbola, pinpoint its center, and determine the coordinates of its vertices. Let's dive into the fascinating world of hyperbolas and unlock the secrets hidden within their equations.

Demystifying the Hyperbola Equation

To begin our exploration, let's first examine the general equation of a hyperbola: $-4x^2 + y^2 + 16x + 12y + 16 = 0$. This equation, while seemingly complex, holds the key to unlocking the hyperbola's characteristics. Our initial goal is to transform this equation into a more manageable standard form, which will readily reveal the center and vertices. The process of completing the square will be our guiding light in this transformation.

Completing the Square: A Pathway to Clarity

Completing the square is a powerful algebraic technique that allows us to rewrite quadratic expressions in a more insightful form. In the context of hyperbolas, this technique helps us to rewrite the general equation into the standard form, which directly reveals the center and the distances to the vertices along the major axis. To achieve this transformation, we will systematically manipulate the given equation, grouping the $x$ terms and the $y$ terms, and then completing the square for each variable separately. This meticulous process will ultimately unveil the hyperbola's essential features.

Let's embark on the journey of completing the square for our hyperbola equation. First, we group the $x$ terms and the $y$ terms together:

$(-4x^2 + 16x) + (y^2 + 12y) + 16 = 0$

Next, we factor out the coefficients of the squared terms. For the $x$ terms, we factor out a $-4$, and for the $y$ terms, the coefficient is already $1$:

$-4(x^2 - 4x) + (y^2 + 12y) + 16 = 0$

Now comes the core of the completing the square technique. We take half of the coefficient of the linear term (the term with $x$ or $y$), square it, and add it inside the parentheses. However, we must also account for the factor we factored out earlier. For the $x$ terms, half of $-4$ is $-2$, and $(-2)^2 = 4$. So, we add $4$ inside the parentheses, but we're actually adding $-4 imes 4 = -16$ to the left side of the equation. For the $y$ terms, half of $12$ is $6$, and $6^2 = 36$. So, we add $36$ inside the parentheses. Our equation now looks like this:

$-4(x^2 - 4x + 4) + (y^2 + 12y + 36) + 16 = -16 + 36$

We can now rewrite the expressions inside the parentheses as perfect squares:

$-4(x - 2)^2 + (y + 6)^2 + 16 = 20$

Finally, we subtract $16$ from both sides and divide by $4$ to get the standard form of the hyperbola equation:

$-4(x - 2)^2 + (y + 6)^2 = 4$

Dividing both sides by 4, we get:

$-(x - 2)^2 + \frac{(y + 6)^2}{4} = 1$

Rearranging the terms, we have the standard form equation:

$\frac{(y + 6)^2}{4} - (x - 2)^2 = 1$

This standard form unveils the hyperbola's essential characteristics, paving the way for us to pinpoint its center and vertices.

Unveiling the Center of the Hyperbola

The standard form equation of a hyperbola, which we derived in the previous section, is given by:

$\frac{(y + 6)^2}{4} - (x - 2)^2 = 1$

This equation takes the general form:

$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$

where $(h, k)$ represents the coordinates of the hyperbola's center. By carefully comparing our derived equation with the general form, we can readily identify the center of our specific hyperbola.

In our case, we observe that $h = 2$ and $k = -6$. Therefore, the center of the hyperbola is located at the point $(2, -6)$. This point serves as the central reference point for the hyperbola, and all other key features, such as the vertices and foci, are defined relative to this center. Understanding the center is paramount for accurately sketching and analyzing the hyperbola's behavior.

Pinpointing the Vertices of the Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its major axis. The major axis is the line segment that passes through the center of the hyperbola and contains the foci. In the standard form equation, the major axis is determined by the variable with the positive coefficient. In our equation, $\frac{(y + 6)^2}{4} - (x - 2)^2 = 1$, the $y$ term has the positive coefficient, indicating that the major axis is vertical.

The distance from the center to each vertex is given by the value of $a$. In our equation, $a^2 = 4$, so $a = 2$. Since the major axis is vertical, the vertices are located $a$ units above and below the center. The center of our hyperbola is $(2, -6)$. To find the vertices, we add and subtract $a = 2$ from the $y$-coordinate of the center:

• Vertex 1: $(2, -6 + 2) = (2, -4)$
• Vertex 2: $(2, -6 - 2) = (2, -8)$

Therefore, the vertices of the hyperbola are located at the points $(2, -4)$ and $(2, -8)$. These vertices play a crucial role in defining the shape and orientation of the hyperbola. They serve as the endpoints of the transverse axis, which is the line segment that connects the two branches of the hyperbola.

Conclusion: Mastering the Hyperbola's Essence

In this comprehensive guide, we have successfully navigated the intricacies of a hyperbola equation, extracting its essential features. We began by transforming the general equation into the standard form through the meticulous process of completing the square. This transformation paved the way for us to readily identify the hyperbola's center, which we found to be located at $(2, -6)$. We then delved into the concept of vertices, understanding their relationship to the major axis and the value of $a$. By applying this knowledge, we pinpointed the vertices of our hyperbola at $(2, -4)$ and $(2, -8)$.

By mastering these techniques, you have equipped yourself with the tools to analyze and understand hyperbolas effectively. The ability to determine the center and vertices is fundamental to sketching the hyperbola accurately and interpreting its properties. As you continue your exploration of conic sections, remember the principles we have discussed here, and you will be well-prepared to tackle even more complex challenges.

In summary, the equation $-4x^2 + y^2 + 16x + 12y + 16 = 0$ represents a hyperbola. Through the process of completing the square, we transformed this equation into the standard form, which revealed the hyperbola's center at $(2, -6)$ and its vertices at $(2, -4)$ and $(2, -8)$. These key features provide a comprehensive understanding of the hyperbola's shape, orientation, and position in the coordinate plane. This detailed exploration exemplifies the power of algebraic manipulation in uncovering the geometric essence of conic sections.

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