Finding The Equation Of A Hyperbola With Co-vertices (0,7), (0,-7) And Transverse Axis 12

Determining the equation of a hyperbola requires understanding its key features: the co-vertices and the transverse axis. In this article, we will delve into how to find the equation of a hyperbola when given its co-vertices and the length of its transverse axis. Specifically, we will address the scenario where the co-vertices are at $(0, 7)$ and $(0, -7)$, and the transverse axis has a length of 12 units. This exploration will not only provide a step-by-step solution but also enhance your comprehension of hyperbolas and their equations.

Understanding Hyperbolas

Before diving into the specifics, let's establish a solid understanding of hyperbolas. A hyperbola is a type of conic section formed by the intersection of a plane and a double cone. It is defined as the set of all points in a plane such that the absolute difference of the distances from two fixed points (called foci) is constant. This definition leads to the distinctive two-branch shape of a hyperbola.

Key components of a hyperbola include:

• Center: The midpoint of the segment connecting the foci.
• Foci: The two fixed points used in the definition of the hyperbola.
• Vertices: The points on the hyperbola closest to the center; they lie on the transverse axis.
• Co-vertices: The points on the hyperbola that lie on the conjugate axis, perpendicular to the transverse axis.
• Transverse Axis: The axis that passes through the foci and vertices. Its length is $2a$, where $a$ is the distance from the center to a vertex.
• Conjugate Axis: The axis perpendicular to the transverse axis, passing through the center. Its length is $2b$, where $b$ is the distance from the center to a co-vertex.
• Asymptotes: Lines that the hyperbola approaches as it extends to infinity. These lines intersect at the center of the hyperbola and are crucial for sketching the hyperbola.

The standard form equation of a hyperbola centered at $(h, k)$ depends on whether the transverse axis is horizontal or vertical:

• Horizontal Transverse Axis: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
• Vertical Transverse Axis: $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

The relationship between $a$, $b$, and the distance from the center to a focus $c$ is given by $c^2 = a^2 + b^2$. This relationship is essential for finding the foci of the hyperbola.

Determining the Hyperbola's Equation

Now, let's tackle the problem at hand: finding the equation of a hyperbola with co-vertices at $(0, 7)$ and $(0, -7)$ and a transverse axis of 12 units. By systematically analyzing these pieces of information, we can construct the hyperbola's equation step by step. The co-vertices play a crucial role in determining the orientation and shape of the hyperbola, while the length of the transverse axis helps us find the value of 'a'.

Step 1: Identify the Center

The center of the hyperbola is the midpoint of the segment connecting the co-vertices. Given the co-vertices at $(0, 7)$ and $(0, -7)$, the center can be found using the midpoint formula:

$$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Substituting the coordinates of the co-vertices:

$$\left(\frac{0 + 0}{2}, \frac{7 + (-7)}{2}\right) = (0, 0)$$

Thus, the center of the hyperbola is at the origin, $(0, 0)$.

Step 2: Determine the Orientation

Since the co-vertices are vertically aligned at $(0, 7)$ and $(0, -7)$, the conjugate axis is vertical. Consequently, the transverse axis must be horizontal. This indicates that the hyperbola opens left and right, and its standard form equation will be:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Step 3: Find the Value of a

The transverse axis has a length of 12 units. The length of the transverse axis is equal to $2a$, where $a$ is the distance from the center to a vertex. Therefore:

$$2a = 12$$

$$a = 6$$

So, $a^2 = 6^2 = 36$.

Step 4: Find the Value of b

The co-vertices are located at $(0, 7)$ and $(0, -7)$. The distance from the center $(0, 0)$ to a co-vertex is $b$. Therefore, $b = 7$, and $b^2 = 7^2 = 49$.

Step 5: Write the Equation

Now that we have the values of $a^2$ and $b^2$, we can write the equation of the hyperbola:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Substituting $a^2 = 36$ and $b^2 = 49$:

$$\frac{x^2}{36} - \frac{y^2}{49} = 1$$

Therefore, the equation of the hyperbola with co-vertices at $(0, 7)$ and $(0, -7)$ and a transverse axis of 12 units is $\frac{x^2}{36} - \frac{y^2}{49} = 1$.

Additional Insights into Hyperbolas

Beyond finding the equation, understanding other aspects of the hyperbola can provide a more comprehensive view. Let's explore how to find the foci and the equations of the asymptotes for the hyperbola we just defined.

Finding the Foci

The foci are points located along the transverse axis, equidistant from the center. The distance from the center to each focus is denoted by $c$, and it is related to $a$ and $b$ by the equation:

$$c^2 = a^2 + b^2$$

In our case, $a^2 = 36$ and $b^2 = 49$, so:

$$c^2 = 36 + 49 = 85$$

$$c = \sqrt{85}$$

Since the transverse axis is horizontal and the center is at $(0, 0)$, the foci are located at $(\pm c, 0)$, which means the foci are at $(-\sqrt{85}, 0)$ and $(\sqrt{85}, 0)$.

Determining the Asymptotes

Asymptotes are lines that the hyperbola approaches as $x$ and $y$ become very large. They intersect at the center of the hyperbola and provide a framework for sketching the hyperbola. The equations of the asymptotes for a hyperbola centered at the origin are given by:

$$y = \pm \frac{b}{a}x$$

In our case, $a = 6$ and $b = 7$, so the equations of the asymptotes are:

$$y = \pm \frac{7}{6}x$$

Thus, the asymptotes are $y = \frac{7}{6}x$ and $y = -\frac{7}{6}x$.

Conclusion

In this article, we have successfully determined the equation of a hyperbola given its co-vertices and the length of its transverse axis. We began by understanding the basic definition and components of a hyperbola, including the center, foci, vertices, co-vertices, transverse axis, and conjugate axis. We then applied this knowledge to find the equation of a hyperbola with co-vertices at $(0, 7)$ and $(0, -7)$ and a transverse axis of 12 units. By identifying the center, determining the orientation, and calculating the values of $a$ and $b$, we derived the equation $\frac{x^2}{36} - \frac{y^2}{49} = 1$. Additionally, we explored how to find the foci and the equations of the asymptotes, further enhancing our understanding of hyperbolas.

Mastering the concepts of hyperbolas and their equations is crucial for various applications in mathematics, physics, and engineering. By practicing these techniques and understanding the underlying principles, you can confidently solve problems involving hyperbolas and their properties. Whether you're a student learning conic sections or a professional applying mathematical concepts in your field, a strong grasp of hyperbolas will undoubtedly be a valuable asset.

This comprehensive guide not only provides a solution to the specific problem but also offers a deep dive into the world of hyperbolas, ensuring that you are well-equipped to tackle similar challenges in the future. Remember, the key to mastering mathematics lies in understanding the fundamentals and practicing consistently. Keep exploring, keep learning, and you'll find that even the most complex concepts become clear with time and effort.