Find the equation of Hyperbola given vertices and asymptote

Prof D
27 Apr 202108:01
EducationalLearning
32 Likes 10 Comments

TLDRThis educational video tutorial guides viewers on deriving the standard form equation of a hyperbola when given its vertices and the equation of its asymptote. Two examples are provided: the first explains how to find the equation for a horizontally opening hyperbola with vertices at (-6, 0) and (6, 0), and an asymptote of y = (4/3)x, resulting in the equation x²/36 - y²/64 = 1. The second example demonstrates the process for a vertically opening hyperbola with vertices at (0, 4) and (0, -4), and an asymptote of y = (2/3)x, leading to the equation y²/16 - x²/36 = 1. The video is a helpful resource for understanding the geometric properties and mathematical formulation of hyperbolas.

Takeaways
  • 📚 The video is a math tutorial focused on finding the standard form of the equation for a hyperbola given its vertices and the equation of the asymptote.
  • 📐 The first example involves a hyperbola with vertices at (-6, 0) and (6, 0) and an asymptote equation of y = (4/3)x.
  • 📈 The standard form for a hyperbola with a horizontal transverse axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
  • 🔍 The distance from the center to a vertex (a) is found to be 6 units for the first example.
  • 🧐 The relationship between the asymptote and the hyperbola's axes is used to find the value of b, resulting in b = 8 for the first example.
  • 📝 The final equation for the first hyperbola is \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \).
  • 📍 The second example has vertices at (0, 4) and (0, -4) with an asymptote equation of y = (2/3)x.
  • 📊 For a hyperbola with a vertical transverse axis, the standard form is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
  • 📏 The distance from the center to a vertex (a) is 4 units for the second example.
  • 🔢 Using the asymptote equation, the value of b is calculated to be 6 for the second example.
  • 📑 The final equation for the second hyperbola is \( \frac{y^2}{16} - \frac{x^2}{36} = 1 \).
  • 👨‍🏫 The presenter is Prof D, who encourages viewers to ask questions or seek clarifications in the comment section.
Q & A

  • What is the main topic of the video?

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

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

    -In example one, the vertices are at (-6, 0) and (6, 0).

  • What is the equation of the asymptote in example one?

    -The equation of the asymptote in example one is y = (4/3)x.

  • Why does the hyperbola in example one open horizontally?

    -The hyperbola opens horizontally because its vertices form a horizontal line, indicating a horizontal transverse axis.

  • What is the general form of the equation for a horizontally opening hyperbola?

    -The general form of the equation for a horizontally opening hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

  • How is the value of 'a' determined in example one?

    -In example one, 'a' is determined by the distance between the center and a vertex, which is 6 units, so 'a' equals 6.

  • How is the value of 'b' calculated in example one?

    -In example one, 'b' is calculated using the equation of the asymptote, where \( \frac{4}{3} = \frac{b}{6} \), leading to \( b = 8 \) after cross-multiplication and solving.

  • What is the standard form of the hyperbola's equation in example one?

    -The standard form of the hyperbola's equation in example one is \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \).

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

    -In example two, the vertices are at (0, 4) and (0, -4).

  • What is the equation of the asymptote in example two?

    -The equation of the asymptote in example two is y = (2/3)x.

  • Why does the hyperbola in example two open vertically?

    -The hyperbola opens vertically because its vertices form a vertical line, indicating a vertical transverse axis.

  • What is the general form of the equation for a vertically opening hyperbola?

    -The general form of the equation for a vertically opening hyperbola is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

  • How is the value of 'a' determined in example two?

    -In example two, 'a' is determined by the distance between the center and a vertex, which is 4 units, so 'a' equals 4.

  • How is the value of 'b' calculated in example two?

    -In example two, 'b' is calculated using the equation of the asymptote, where \( \frac{2}{3} = \frac{a}{b} \) and since 'a' is 4, it leads to \( b = 6 \) after solving.

  • What is the standard form of the hyperbola's equation in example two?

    -The standard form of the hyperbola's equation in example two is \( \frac{y^2}{16} - \frac{x^2}{36} = 1 \).

  • What advice does the instructor give at the end of the video?

    -The instructor advises viewers to leave any questions or clarifications in the comment section if they need further assistance.

Outlines
00:00
📚 Introduction to Finding Hyperbola Equations

This paragraph introduces a video tutorial on how to derive the standard form equation of a hyperbola given its vertices and the equation of its asymptote. The video provides a step-by-step guide, starting with an example of a hyperbola with vertices at (-6, 0) and (6, 0) and an asymptote equation of y = (4/3)x. It explains the concept of the transverse axis and how it dictates the orientation of the hyperbola, leading to the selection of the appropriate equation format for a horizontally opening hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

05:01
📐 Calculating Hyperbola Parameters from Vertices and Asymptote

The second paragraph delves into the process of calculating the parameters 'a' and 'b' for the hyperbola's equation using the vertices and the asymptote. It first establishes 'a' as the distance from the center to a vertex, which is determined to be 6 units in this case. Then, using the asymptote equation, the relationship between 'a' and 'b' is found by setting up a proportion, leading to the calculation of 'b' as 8 units. The standard form of the hyperbola's equation is then presented as \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \). The paragraph also introduces a second example with different vertices and an asymptote, setting the stage for another calculation of 'a' and 'b', but the explanation for this example is cut off, leaving the process incomplete.

Mindmap
Keywords
💡Hyperbola
A hyperbola is a type of conic section that resembles two mirror-imaged parabolas opening away from each other. In the context of the video, the theme revolves around finding the standard form equation of a hyperbola given its vertices and the equation of the asymptote. The hyperbola is a key concept as it is the main geometric figure being discussed and analyzed.
💡Standard form
The standard form of an equation is a specific way of writing algebraic expressions that allows for easy identification of the properties of the graph it represents. In the video, the standard form is essential for expressing the hyperbola's equation, which is critical for understanding its geometric characteristics and for solving related mathematical problems.
💡Vertices
Vertices are the points where the hyperbola turns, and they are crucial in determining the shape and position of the hyperbola. In the script, vertices are given as coordinates, such as (-6,0) and (6,0), and they help in identifying the orientation and size of the hyperbola.
💡Asymptote
An asymptote is a line that a curve approaches but never meets. For hyperbolas, asymptotes are important because they provide insight into the behavior of the hyperbola at infinity. In the video, the equation of the asymptote, such as 'y = 4x/3', is used to find the relationship between 'a' and 'b' in the hyperbola's equation.
💡Transverse axis
The transverse axis is the line segment that passes through the vertices of the hyperbola and is perpendicular to the conjugate axis. It indicates the direction in which the hyperbola opens. The video explains that if the vertices form a horizontal line, the hyperbola opens horizontally, and if they form a vertical line, it opens vertically.
💡Equation
In the context of the video, an equation is a mathematical statement that shows the relationship between different quantities. Specifically, it refers to the equations of the hyperbola and its asymptote. The script uses equations to derive the standard form of the hyperbola's equation.
💡Center
The center of a hyperbola is the midpoint between its vertices and is a key point in determining the hyperbola's position. In the video, the center is used as a reference to calculate distances to the vertices and to establish the hyperbola's orientation.
💡Distance
Distance in the video refers to the length between two points, specifically the distance from the center to a vertex, which is denoted as 'a'. This distance is essential for determining the shape of the hyperbola and is used in the calculation of 'b'.
💡Substitution
Substitution is a method used in algebra where the value of one expression is replaced with another to simplify the equation. In the video, substitution is used to solve for 'b' by replacing 'a' with its calculated value in the equation derived from the asymptote.
💡Cross multiplication
Cross multiplication is a technique used to solve proportions or to clear fractions in equations. The script mentions cross multiplication when solving for 'b' by multiplying both sides of the proportion '4/3 = b/a' to find the value of 'b'.
💡Formula
A formula in mathematics is a concise way of expressing information symbolically as a mathematical sentence. In the video, the formula refers to the standard form equations of a hyperbola, which are derived from the given vertices and asymptote equations.
Highlights

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

Example 1: Hyperbola with vertices at (-6, 0) and (6, 0) and asymptote y = (4/3)x.

Explanation of the transverse axis and its relation to the orientation of the hyperbola.

The standard form of a hyperbola with a horizontal transverse axis: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

Calculation of the distance 'a' from the center to the vertex, which is 6 units.

Using the asymptote equation to find the relationship between 'a' and 'b'.

Derivation of 'b' value using the ratio \( \frac{4}{3} = \frac{b}{a} \) where 'a' is 6.

Final equation for Example 1: \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \).

Introduction to Example 2 with vertices at (0, 4), (0, -4) and asymptote y = (2/3)x.

Hyperbola with a vertical transverse axis and the corresponding standard form.

Calculation of 'a' as the distance between the center and vertex, which is 4 units.

Applying the asymptote equation to find 'b' for a vertically opening hyperbola.

Determining 'b' using the relationship \( \frac{2}{3} = \frac{a}{b} \) with 'a' being 4.

Final equation for Example 2: \( \frac{y^2}{16} - \frac{x^2}{36} = 1 \).

Conclusion of the video with a prompt for questions or clarifications.

Sign off by Prof D, indicating the end of the video.

Transcripts

Browse More Related Video

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

How to find the equation of an ellipse given foci and vertices | @ProfD

STANDARD EQUATION OF PARABOLA TO ITS GENERAL FORM

GENERAL EQUATION OF THE HYPERBOLA IN STANDARD FORM || PRE-CALCULUS

How To Find The Equation of a Secant Line

Area Between y=x^3 and y=3x-2 | MIT 18.01SC Single Variable Calculus, Fall 2010

Rate This

5.0 / 5 (0 votes)

Thanks for rating:

Related Tags
Hyperbola EquationsVerticesAsymptotesMath TutorialEducational ContentGeometry LessonsStandard FormTransverse AxisConic SectionsAlgebraic GeometryMathematics Education