A Vase Can Be Modeled Using The Equation:${ \frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1 }$and The $x$-axis, For $0 \leq Y \leq 20$, Where The Measurements Are In Inches.Using The Graph, What Is The Distance Across The

Introduction

In mathematics, equations are used to describe various shapes and objects. One such equation is the one used to model a vase, which is given by $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$. This equation represents a hyperbola, a type of curve that has two branches and is symmetrical about its center. In this article, we will explore the graph of this equation and use it to find the distance across the vase.

Understanding the Equation

The given equation is a hyperbola with its center at $(0, 4)$. The equation can be rewritten as $\frac{x^2}{(2.5)^2} - \frac{(y-4)^2}{(6.5)^2} = 1$, which makes it easier to understand the shape of the curve. The values $2.5$ and $6.5$ represent the distances from the center to the vertices of the hyperbola along the $x$-axis and $y$-axis, respectively.

Graphing the Equation

To graph the equation, we can start by plotting the center of the hyperbola at $(0, 4)$. Then, we can plot the vertices of the hyperbola, which are located at $(2.5, 4)$ and $(-2.5, 4)$. The asymptotes of the hyperbola are given by the equations $y = 4 + \frac{6.5}{2.5}(x-0)$ and $y = 4 - \frac{6.5}{2.5}(x-0)$. These asymptotes intersect at the center of the hyperbola and help to define the shape of the curve.

Finding the Distance Across the Vase

To find the distance across the vase, we need to find the distance between the two vertices of the hyperbola. Since the vertices are located at $(2.5, 4)$ and $(-2.5, 4)$, the distance between them can be found using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Plugging in the values, we get:

$$d = \sqrt{(-2.5 - 2.5)^2 + (4 - 4)^2}$$

Simplifying the expression, we get:

$$d = \sqrt{(-5)^2 + 0^2}$$

$$d = \sqrt{25}$$

$$d = 5$$

Therefore, the distance across the vase is $5$ inches.

Conclusion

In this article, we used the equation $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$ to model a vase. We graphed the equation and used it to find the distance across the vase. The distance was found to be $5$ inches, which is the distance between the two vertices of the hyperbola. This example demonstrates how equations can be used to describe various shapes and objects in mathematics.

Additional Information

• The equation $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$ represents a hyperbola with its center at $(0, 4)$.
• The vertices of the hyperbola are located at $(2.5, 4)$ and $(-2.5, 4)$.
• The asymptotes of the hyperbola are given by the equations $y = 4 + \frac{6.5}{2.5}(x-0)$ and $y = 4 - \frac{6.5}{2.5}(x-0)$.
• The distance across the vase is $5$ inches.

References

• [1] "Hyperbola". Math Open Reference. Retrieved 2023-02-20.
• [2] "Equation of a Hyperbola". Mathway. Retrieved 2023-02-20.
• [3] "Graphing a Hyperbola". Khan Academy. Retrieved 2023-02-20.
  

Q&A: Understanding the Equation and Graphing the Vase

Q: What is the equation used to model the vase?

A: The equation used to model the vase is $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$. This equation represents a hyperbola, a type of curve that has two branches and is symmetrical about its center.

Q: What is the center of the hyperbola?

A: The center of the hyperbola is located at $(0, 4)$. This is the point around which the hyperbola is symmetrical.

Q: What are the vertices of the hyperbola?

A: The vertices of the hyperbola are located at $(2.5, 4)$ and $(-2.5, 4)$. These points are the endpoints of the hyperbola and are used to define its shape.

Q: What are the asymptotes of the hyperbola?

A: The asymptotes of the hyperbola are given by the equations $y = 4 + \frac{6.5}{2.5}(x-0)$ and $y = 4 - \frac{6.5}{2.5}(x-0)$. These lines intersect at the center of the hyperbola and help to define its shape.

Q: How do you find the distance across the vase?

A: To find the distance across the vase, you need to find the distance between the two vertices of the hyperbola. This can be done using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Q: What is the distance across the vase?

A: The distance across the vase is $5$ inches. This is the distance between the two vertices of the hyperbola.

Q: What is the significance of the equation in real-world applications?

A: The equation $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$ can be used to model various shapes and objects in real-world applications, such as architecture, engineering, and design.

Q: How can you modify the equation to model different shapes?

A: To modify the equation to model different shapes, you can change the values of the coefficients in the equation. For example, you can change the value of $6.25$ to $4$ to model a different shape.

Q: What are some common applications of hyperbolas in mathematics?

A: Hyperbolas have many applications in mathematics, including:

• Modeling the shape of a vase or a container
• Describing the motion of an object under the influence of gravity
• Defining the shape of a mirror or a lens
• Modeling the shape of a satellite or a spacecraft

Conclusion

In this Q&A article, we have explored the equation $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$ and its applications in modeling a vase. We have also discussed the center, vertices, and asymptotes of the hyperbola, as well as the distance across the vase. We hope that this article has provided a clear understanding of the equation and its significance in real-world applications.

Additional Information

• The equation $\frac{x^2}{6.25} - \frac{(y-4)^2}{42.25} = 1$ represents a hyperbola with its center at $(0, 4)$.
• The vertices of the hyperbola are located at $(2.5, 4)$ and $(-2.5, 4)$.
• The asymptotes of the hyperbola are given by the equations $y = 4 + \frac{6.5}{2.5}(x-0)$ and $y = 4 - \frac{6.5}{2.5}(x-0)$.
• The distance across the vase is $5$ inches.

References

• [1] "Hyperbola". Math Open Reference. Retrieved 2023-02-20.
• [2] "Equation of a Hyperbola". Mathway. Retrieved 2023-02-20.
• [3] "Graphing a Hyperbola". Khan Academy. Retrieved 2023-02-20.