Finding The Center And Vertices Of A Hyperbola

Finding the center and vertices of a hyperbola can seem daunting, but with a systematic approach, it becomes a manageable task. In this article, we'll break down the process step-by-step, using the given equation as an example. We will explore the techniques to transform the general equation of a hyperbola into its standard form, which allows us to easily identify its key features, including the center and vertices. This comprehensive guide aims to equip you with the knowledge and skills to confidently tackle similar problems in the future. Understanding these concepts is crucial for various applications in mathematics, physics, and engineering, where hyperbolas play a significant role in describing phenomena like the paths of comets or the shape of cooling towers.

1. Understanding the Hyperbola Equation

Before we dive into the solution, let's understand the hyperbola equation we're working with:

$$25x^2 - 16y^2 + 50x + 128y - 631 = 0$$

This equation is in the general form of a hyperbola. To find the center and vertices, we need to convert it into the standard form. The standard form of a hyperbola equation depends on whether the hyperbola opens horizontally or vertically. A hyperbola is a conic section formed by the intersection of a double cone with a plane. It consists of two separate curves, or branches, that are mirror images of each other. The standard form provides critical information about the hyperbola's orientation, center, vertices, foci, and asymptotes. Recognizing the general form and knowing how to transform it is the first step toward analyzing and understanding the properties of a hyperbola. The standard forms allow for easy extraction of the hyperbola's key parameters, which are necessary for graphing and problem-solving.

2. Completing the Square

The key technique to transform the general form into the standard form is completing the square. This involves grouping the x terms and the y terms, and then manipulating each group to form perfect square trinomials. Completing the square is a fundamental algebraic technique used to rewrite quadratic expressions. It involves adding and subtracting a specific constant to create a perfect square trinomial, which can then be factored into the square of a binomial. This process is crucial for transforming the general equation of a conic section (such as a hyperbola, ellipse, or parabola) into its standard form. By completing the square for both the x and y terms, we can rewrite the given equation in a form that readily reveals the hyperbola's center, vertices, and other essential properties. This method is not only applicable to hyperbolas but also serves as a powerful tool in various algebraic manipulations and problem-solving scenarios.

2.1 Grouping x and y terms

First, let's group the x terms and y terms together:

$$(25x^2 + 50x) + (-16y^2 + 128y) = 631$$

Grouping like terms is a preliminary step in simplifying and rearranging equations. By bringing together terms with the same variable, we set the stage for further algebraic manipulations. In this case, grouping the x and y terms allows us to focus on each variable separately when completing the square. This strategic separation simplifies the process and reduces the chances of errors. It's a common practice in algebra to organize equations by grouping similar terms, as it facilitates a clearer understanding of the equation's structure and the relationships between variables. This step is crucial for applying techniques such as completing the square, which require a focused approach on individual variable expressions.

2.2 Factoring out coefficients

Next, we factor out the coefficients of the squared terms:

$$25(x^2 + 2x) - 16(y^2 - 8y) = 631$$

Factoring out the coefficients of the squared terms is an essential step in the process of completing the square. By factoring out the leading coefficients (25 for the x terms and -16 for the y terms), we ensure that the quadratic expressions inside the parentheses have a leading coefficient of 1. This is a prerequisite for successfully completing the square. When the leading coefficient is 1, we can easily determine the constant term needed to create a perfect square trinomial. Factoring out the coefficients also simplifies the subsequent steps and allows for a more straightforward application of the completing the square technique. This step demonstrates a key algebraic principle: simplifying expressions to reveal their underlying structure and make them easier to manipulate.

2.3 Completing the square for x

To complete the square for the x terms, we take half of the coefficient of the x term (which is 2), square it (1), and add it inside the parenthesis. Remember to add 25 times this value to the right side to balance the equation:

$$25(x^2 + 2x + 1) - 16(y^2 - 8y) = 631 + 25(1)$$

Completing the square involves transforming a quadratic expression into a perfect square trinomial, which can then be factored into the square of a binomial. For the x terms, we take half of the coefficient of the x term (which is 2), square it (resulting in 1), and add it inside the parentheses. This addition creates the perfect square trinomial x² + 2x + 1. However, since we've added 1 inside the parentheses that are multiplied by 25, we must add 25(1) to the right side of the equation to maintain balance. This step is crucial for rewriting the equation in standard form. By creating perfect square trinomials, we can factor them into binomial squares, which simplifies the equation and allows us to identify the center and other key features of the hyperbola. This technique is a fundamental tool in algebraic manipulations and is widely used in solving various mathematical problems.

2.4 Completing the square for y

Similarly, for the y terms, we take half of the coefficient of the y term (which is -8), square it (16), and add it inside the parenthesis. Remember to subtract 16 times this value from the right side to balance the equation (because of the -16 outside the parenthesis):

$$25(x^2 + 2x + 1) - 16(y^2 - 8y + 16) = 631 + 25(1) - 16(16)$$

Completing the square for the y terms follows the same principle as for the x terms but with careful attention to the coefficients. We take half of the coefficient of the y term (which is -8), square it (resulting in 16), and add it inside the parentheses. This addition creates the perfect square trinomial y² - 8y + 16. However, since we've added 16 inside the parentheses that are multiplied by -16, we must subtract 16(16) from the right side of the equation to maintain balance. This step is critical for correctly rewriting the equation in standard form. The negative sign in front of the 16 requires us to subtract the added term on the right side, which is a common point of error. Accurate application of this step ensures that the equation remains balanced and the hyperbola's properties can be correctly determined.

3. Rewriting in Standard Form

Now, we rewrite the equation by factoring the perfect square trinomials and simplifying the right side:

$$25(x + 1)^2 - 16(y - 4)^2 = 631 + 25 - 256$$

$$25(x + 1)^2 - 16(y - 4)^2 = 400$$

Rewriting the equation involves factoring the perfect square trinomials created in the previous steps and simplifying the right side. The trinomial x² + 2x + 1 is factored into (x + 1)², and the trinomial y² - 8y + 16 is factored into (y - 4)². On the right side, we perform the arithmetic operation 631 + 25 - 256, which simplifies to 400. This step is crucial for transforming the equation into a form that closely resembles the standard equation of a hyperbola. By expressing the quadratic expressions as squared binomials and simplifying the constant term, we bring the equation closer to its standard form, which makes it easier to identify the center, vertices, and other essential parameters of the hyperbola. This algebraic manipulation is a key step in understanding and analyzing conic sections.

3.1 Dividing to get 1 on the right side

Divide both sides by 400 to get the standard form:

rac{(x + 1)^2}{16} - rac{(y - 4)^2}{25} = 1

Dividing both sides of the equation by 400 is the final step in transforming the equation into the standard form of a hyperbola. This division ensures that the constant term on the right side of the equation becomes 1, which is a defining characteristic of the standard form. The resulting equation, ((x + 1)²)/16 - ((y - 4)²)/25 = 1, clearly displays the structure of a hyperbola. From this form, we can easily identify the center, the lengths of the semi-major and semi-minor axes, and the orientation of the hyperbola. This step is crucial for analyzing the hyperbola's properties and for graphing it accurately. The standard form provides a clear and concise representation of the hyperbola's key features, making it easier to work with and understand its characteristics.

4. Identifying the Center

The center of the hyperbola is given by the coordinates (h, k), where h and k are the values that make the expressions inside the squared terms zero. In our equation:

rac{(x + 1)^2}{16} - rac{(y - 4)^2}{25} = 1

• h = -1 (because x + 1 = 0 implies x = -1)
• k = 4 (because y - 4 = 0 implies y = 4)

Thus, the center is (-1, 4).

Identifying the center of the hyperbola is a straightforward process once the equation is in standard form. The center's coordinates, represented as (h, k), can be directly extracted from the equation. In the standard form, the x-coordinate of the center (h) is the value that makes the expression inside the squared term involving x equal to zero, and the y-coordinate (k) is the value that makes the expression inside the squared term involving y equal to zero. In the given equation, ((x + 1)²)/16 - ((y - 4)²)/25 = 1, setting x + 1 = 0 gives h = -1, and setting y - 4 = 0 gives k = 4. Therefore, the center of the hyperbola is located at the point (-1, 4). This central point serves as a reference for determining other key features of the hyperbola, such as the vertices, foci, and asymptotes. The center is a fundamental characteristic of a hyperbola and is essential for its graphical representation and analysis.

5. Finding the Vertices

Since the x² term is positive, the hyperbola opens horizontally. The distance from the center to each vertex is a, where a² is the denominator under the x² term. In our equation, a² = 16, so a = 4.

To find the vertices, we move a units to the left and right from the center:

• Vertex 1: (-1 - 4, 4) = (-5, 4)
• Vertex 2: (-1 + 4, 4) = (3, 4)

Therefore, the vertices are (-5, 4) and (3, 4).

Finding the vertices of a hyperbola is a crucial step in understanding its shape and orientation. The vertices are the points where the hyperbola intersects its transverse axis, which is the axis that passes through the center and the foci. To determine the vertices, we first identify whether the hyperbola opens horizontally or vertically. This is indicated by the term that is positive in the standard form equation. If the x² term is positive, the hyperbola opens horizontally, and if the y² term is positive, it opens vertically. In this case, since the x² term is positive, the hyperbola opens horizontally. The distance from the center to each vertex is represented by the variable a, which is the square root of the denominator under the positive squared term in the standard form equation. For the given equation, a² = 16, so a = 4. To find the coordinates of the vertices, we move a units to the left and right from the center along the transverse axis. Thus, the vertices are located at (-1 - 4, 4) = (-5, 4) and (-1 + 4, 4) = (3, 4). These points are essential for sketching the hyperbola and for understanding its geometric properties. The vertices, along with the center, provide a fundamental framework for analyzing the hyperbola's characteristics.

6. Summary

For the hyperbola given by the equation $25x^2 - 16y^2 + 50x + 128y - 631 = 0$:

• Center: (-1, 4)
• Vertices: (-5, 4) and (3, 4)

This step-by-step guide has demonstrated how to find the center and vertices of a hyperbola by completing the square and transforming the general equation into standard form. By understanding the process of completing the square and recognizing the standard form of a hyperbola's equation, one can confidently determine these key features. The center and vertices provide essential information about the hyperbola's position, orientation, and shape. This knowledge is not only valuable for mathematical problem-solving but also for applications in various fields, such as physics and engineering, where hyperbolas are used to model different phenomena. Mastering the techniques outlined in this guide provides a solid foundation for further exploration of conic sections and their properties. The systematic approach presented here can be applied to any hyperbola equation, making it a versatile tool for mathematical analysis.

7. Practice Problems

To solidify your understanding, try applying these steps to other hyperbola equations. You can also explore finding the foci and asymptotes of hyperbolas, which are other key characteristics. Practice is key to mastering any mathematical concept, and working through additional problems will help you become more comfortable and confident in your ability to analyze hyperbolas.

By tackling a variety of problems, you'll develop a deeper understanding of the underlying principles and gain proficiency in applying the techniques discussed in this guide. Exploring additional characteristics, such as the foci and asymptotes, will further enhance your knowledge of hyperbolas and their properties. The more you practice, the better you'll become at recognizing patterns, identifying key information, and solving problems efficiently. This hands-on experience is invaluable for building a strong foundation in mathematics and for applying these concepts in real-world scenarios. Remember, each problem you solve brings you closer to mastery.