How to find the equation of Hyperbola given vertex, and the equation of asymptote

Prof D
4 May 202106:50
EducationalLearning
32 Likes 10 Comments

TLDRIn this educational video, Prof D demonstrates how to derive the equation of a hyperbola when given its vertices and the equation of its asymptote. The example provided has vertices at (-4, 2) and (2, 12), with an asymptote equation y = -2x + 8. The video explains that since the vertices form a vertical line, the hyperbola opens vertically. Using the distance formula and the slope of the asymptote, the video calculates the values of 'a' and 'b', and then constructs the standard form of the hyperbola's equation. The final equation is simplified and presented, offering viewers a clear understanding of the process.

Takeaways
  • 📚 The video is a tutorial on finding the equation of a hyperbola given its vertices and the equation of its asymptote.
  • 📍 The example provided has vertices at (-4, 2) and (12, 2), indicating the hyperbola opens vertically.
  • 📐 The center of the hyperbola is calculated to be at (2, 4), which is the midpoint between the vertices.
  • 🔍 The value of 'a' is determined to be 8 units, which is the distance from the center to a vertex.
  • 📈 The slope of the asymptote is given as -2, which is used to find the value of 'b'.
  • 🧩 The value of 'b' is calculated to be -4 by using the slope of the asymptote and cross-multiplication.
  • 📝 The standard form of the hyperbola's equation is derived as (y - k)^2 / a^2 - (x - h)^2 / b^2 = 1.
  • 🔢 The final equation of the hyperbola is simplified to (y - 4)^2 / 64 - (x - 2)^2 / 16 = 1.
  • 👨‍🏫 The presenter is Prof D, who encourages viewers to ask questions or seek clarifications in the comments.
  • 👋 The video concludes with a sign-off from Prof D, indicating the end of the tutorial.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is to demonstrate how to find the equation of a hyperbola given its vertices and the equation of its asymptote.

  • What are the coordinates of the vertices given in the example?

    -The coordinates of the vertices given in the example are (-4, 2) and (12, 2).

  • What does the orientation of the vertices indicate about the hyperbola?

    -The orientation of the vertices indicates that the hyperbola has a vertical transverse axis, meaning its opening is vertical.

  • What is the general form of the equation for a hyperbola with a vertical transverse axis?

    -The general form of the equation for a hyperbola with a vertical transverse axis is \((y - k)^2 / a^2 - (x - h)^2 / b^2 = 1\).

  • What is the center of the hyperbola in the given example?

    -The center of the hyperbola in the given example is (2, 4), which is calculated by averaging the x-coordinates and y-coordinates of the vertices.

  • What is the value of 'a' in the hyperbola equation, and what does it represent?

    -The value of 'a' in the hyperbola equation is 8. It represents the distance between the center of the hyperbola and each vertex along the transverse axis.

  • What is the equation of the asymptote given in the example, and what does it tell us about the hyperbola?

    -The equation of the asymptote given in the example is \(y = -2x + 8\). It tells us the slope of the asymptotes and helps in determining the value of 'b' in the hyperbola equation.

  • How is the value of 'b' in the hyperbola equation determined?

    -The value of 'b' is determined by using the slope of the asymptote and the relationship between 'a' and 'b'. In this case, it is found by solving the equation \(-2b = 8\), which gives \(b = -4\).

  • What is the final equation of the hyperbola in the given example?

    -The final equation of the hyperbola in the given example is \((y - 4)^2 / 64 - (x - 2)^2 / 16 = 1\).

  • What does the video suggest if you have questions or need clarifications?

    -The video suggests that if you have questions or need clarifications, you should leave them in the comment section below.

  • Who is the presenter of the video?

    -The presenter of the video is Prof D.

Outlines
00:00
📚 Welcome and Introduction to Hyperbolas

In this segment, the instructor welcomes the class back to the channel and introduces the topic of finding the equation of a hyperbola given its vertices and asymptote. The example provided includes vertices at (2, -4) and (2, 12), and the asymptote equation y = -2x + 8. The first steps involve identifying the vertices and recognizing the vertical alignment, which determines that the hyperbola's transverse axis and opening are vertical.

05:01
🧮 Calculating Hyperbola Parameters

This section focuses on determining the parameters needed for the hyperbola's equation. The center of the hyperbola is calculated as the midpoint between the vertices, yielding (2, 4). The value of 'a' is identified as the distance from the center to a vertex, which is 8 units. The slope of the asymptote (-2) is used to find the value of 'b' through cross-multiplication, resulting in b = 4. These values are then used to construct the standard form of the hyperbola's equation.

✍️ Finalizing the Hyperbola Equation

The final part of the video simplifies the standard form of the hyperbola's equation using the calculated parameters. The equation y - 4^2 / 64 - (x - 2)^2 / 16 = 1 is derived. The instructor concludes the lesson by encouraging viewers to leave any questions or clarifications in the comments, signing off with a thank you and a promise to see them in the next video.

Mindmap
Keywords
💡Hyperbola
A hyperbola is a type of conic section that resembles two mirror-imaged parabolas. It is defined as the set of all points in a plane, the difference of whose distances from two fixed points (the foci) is constant. In the context of the video, the equation of a hyperbola is being derived from given vertices and the equation of an asymptote, which are key to understanding its shape and orientation.
💡Vertices
Vertices refer to the two points on a hyperbola that are closest to each other and lie on the transverse axis. The script mentions vertices at (-4, 2) and (12, 2), indicating the hyperbola's position and orientation. These points are crucial for determining the center and defining the hyperbola's equation.
💡Asymptote
An asymptote is a line that a curve approaches but does not meet. In the case of a hyperbola, there are two asymptotes that pass through the center and define the 'opening' of the hyperbola. The script provides the equation of an asymptote as y = -2x + 8, which is essential for deriving the hyperbola's equation.
💡Center
The center of a hyperbola is the midpoint between its vertices and is the point from which distances to the foci and vertices are measured. The script identifies the center as being 8 units away from the vertices, which helps in calculating the values of 'a' and 'b' in the hyperbola's equation.
💡Transverse axis
The transverse axis is the line segment that passes through the vertices and the center of the hyperbola. In the script, it is mentioned that the vertices form a vertical line, indicating that the transverse axis is vertical, which is a key characteristic of the hyperbola's orientation.
💡Equation
The equation of a hyperbola is a mathematical formula that describes its shape and position in a coordinate plane. The video script details the process of deriving this equation from given vertices and an asymptote, which is essential for understanding how to represent a hyperbola mathematically.
💡Standard form
The standard form of a hyperbola's equation is a specific format that allows for easy identification of the hyperbola's properties. The script describes transforming the derived equation into the standard form y - k^2 / a^2 - (x - h)^2 / b^2 = 1, which simplifies the analysis of the hyperbola's characteristics.
💡A
In the context of the hyperbola's equation, 'a' represents the distance from the center to a vertex along the transverse axis. The script calculates 'a' as 8 units, which is a critical parameter in determining the hyperbola's shape.
💡B
'B' is the distance from the center to a point on the hyperbola along the conjugate axis, perpendicular to the transverse axis. The script derives 'b' from the slope of the asymptote, which is essential for completing the hyperbola's equation.
💡Slope
Slope is a measure of the steepness of a line and is calculated as the ratio of the vertical change to the horizontal change between two points. In the script, the slope of the asymptote is given as -2, which is used to find the value of 'b' and is crucial for the hyperbola's equation.
Highlights

Introduction to finding the equation of a hyperbola given its vertices and the equation of its asymptote.

Vertices are given as (-4, 2) and (12, 2) indicating the hyperbola opens vertically.

The center of the hyperbola is calculated to be at (4, 4).

The value of 'a' is determined to be 8 units, the distance between the vertex and the center.

The equation of the asymptote is given as y = -2x + 8.

The slope of the asymptote is -2, which is derived from the given equation.

The value of 'b' is calculated using the slope and the center's coordinates.

The equation for 'b' is solved using cross multiplication, resulting in b = -4.

The standard form of the hyperbola equation is introduced.

The equation is simplified into y - 4^2 / 64 - (x - 2)^2 / (-4)^2 = 1.

The final equation of the hyperbola is presented in a simplified form.

An invitation for questions or clarifications is extended to the viewers.

The video concludes with a sign-off from Prof D.

The importance of understanding the hyperbola's orientation based on its vertices is emphasized.

The method of finding the center of the hyperbola using the vertices is explained.

The process of determining the value of 'a' and 'b' for the hyperbola's equation is detailed.

The relationship between the slope of the asymptote and the hyperbola's equation is discussed.

The final equation is derived by substituting the values of 'a', 'b', 'h', and 'k' into the standard form.

Transcripts

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Hyperbola EquationVerticesAsymptotesMathematicsEducationalGeometryConic SectionsGraphingCalculusTutorial