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

WOW MATH

4 Oct 202116:57

EducationalLearning

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TLDRThis video tutorial delves into the general equation of a hyperbola in standard form, represented as Ax^2 + By^2 + Cx + Dy + E = 0. The instructor guides viewers through the process of transforming a given hyperbola equation into its standard form by completing the square for both x and y variables. Through step-by-step examples, the video demonstrates how to simplify equations and achieve the standard form, which is crucial for understanding hyperbolic functions and their geometric representations. The tutorial is designed to help learners master the technique of completing the square and apply it to hyperbola equations, enhancing their mathematical comprehension and problem-solving skills.

Takeaways

📚 The video lesson discusses the general equation of a hyperbola in standard form, which is \( ax^2 + by^2 + cx + dy + e = 0 \).
🔍 The process of converting the general equation to the standard form involves isolating the \( x^2 \) and \( y^2 \) terms and then completing the square for both variables.
📉 The first example provided is \( 4x^2 - y^2 - 24x = 64 \), which is simplified by dividing by the coefficient of \( x^2 \) and completing the square for \( x \) and \( y \).
📶 The standard form of the hyperbola equation is derived by dividing the entire equation by the constant term on the right side, resulting in \( (x-h)^2/a^2 - (y-k)^2/b^2 = 1 \).
🔢 Completing the square involves finding the value that, when added and subtracted to the \( x \) and \( y \) terms, forms a perfect square trinomial.
📐 The coefficient of the linear term in \( x \) or \( y \) is divided by 2, squared, and added to the equation to create the perfect square trinomial.
📈 The second example given is \( 3x^2 - y^2 + 18x + 4y = -35 \), which is also simplified and completed into a standard form hyperbola equation.
📉 The third example, \( x^2 - 4y^2 + 8x - 16y - 36 = 0 \), demonstrates the process of completing the square for both \( x \) and \( y \) and then dividing by the constant term to achieve the standard form.
📝 The final example, \( 16x^2 + 9y^2 + 36y = 64x + 44 \), shows the process of isolating the \( x^2 \) and \( y^2 \) terms, completing the square, and then dividing by the constant term to get the standard form.
👨‍🏫 The instructor emphasizes the importance of dividing by the constant term to simplify the equation to its lowest terms, which is crucial for the standard form of a hyperbola.
👍 The video concludes with a reminder to like, subscribe, and hit the bell button for more video tutorials, indicating the channel's focus on educational content.

Q & A

What is the general equation of a hyperbola in standard form?
-The general equation of a hyperbola in standard form is given by (ax^2 + by^2 + cx + dy + e = 0).
How can you convert the general equation of a hyperbola into its standard form?
-To convert the general equation into the standard form of a hyperbola, you can complete the square for both x and y terms, ensuring the equation resembles (x-h)^2/a^2 - (y-k)^2/b^2 = 1 or (y-k)^2/b^2 - (x-h)^2/a^2 = 1.
What is the first example equation given in the script and how is it transformed into standard form?
-The first example given is (4x^2 - y^2 - 24x = 64). It is transformed by dividing by 4, completing the square for x and y, and then rearranging to get (x-3)^2/25 - y^2/4 = 1.
What is the second example equation, and how is it simplified to the standard form of a hyperbola?
-The second example is (3x^2 - y^2 + 18x + 4y = -35). After completing the square for x and y and rearranging, it simplifies to (x+3)^2/4 - (y-2)^2/12 = 1.
What is the process of completing the square for the x-term in the second example equation?
-For the x-term in the second example, you take the coefficient of the linear term (18x), divide it by 2, and square the result to get 9. This value is then added and subtracted to complete the square, resulting in (x+3)^2.
How do you handle the constant term when completing the square for the y-term in the second example?
-For the y-term, you take the linear coefficient (4y), divide by 2, and square it to get 4. This is then added and subtracted to complete the square, resulting in (y-2)^2.
What is the third example equation provided in the script, and what is its standard form?
-The third example is (x^2 - 4y^2 + 8x - 16y - 36 = 0). After completing the square for both x and y and rearranging, it becomes (x+4)^2/36 - (y+2)^2/9 = 1.
What is the fourth example equation, and how is it simplified to the standard form?
-The fourth example is (16x^2 + 9y^2 + 36y = 64x + 44). After rearranging, completing the square for x and y, and dividing by 144, it simplifies to (x-2)^2/9 + (y+2)^2/16 = 1.
What is the significance of dividing the equation by the constant term after completing the square?
-Dividing the equation by the constant term after completing the square helps to normalize the equation, making it easier to compare with the standard form of a hyperbola and to identify the values of a, b, h, and k.
What is the final step in converting the general equation of a hyperbola into its standard form?
-The final step is to ensure the equation is in the form (x-h)^2/a^2 - (y-k)^2/b^2 = 1 or (y-k)^2/b^2 - (x-h)^2/a^2 = 1, where a and b are positive constants, and h and k are the center coordinates of the hyperbola.

Outlines

00:00

📘 Introduction to the General Equation of Hyperbola

The video begins with an introduction to the general equation of the hyperbola in standard form. It explains how to transform the general equation into its standard form. The example provided involves the equation 4x^2 - y^2 - 24x = 64. The process includes completing the square to convert the equation into the standard form, eventually leading to (x - 3)^2 / 25 - y^2 / 4 = 1.

05:02

📘 Completing the Square for Hyperbola Equations

This section continues with another example involving the equation 3x^2 - y^2 + 18x + 4y = -35. The process of completing the square is applied again to transform the equation into its standard form. Detailed steps show dividing terms, adding constants, and factoring to eventually reach (x + 3)^2 / 4 - (y - 2)^2 / 12 = 1.

10:15

📘 Hyperbola Equation with Mixed Terms

In this part, the video addresses an equation with both x and y terms: x^2 - 4y^2 + 8x - 16y - 36 = 0. The steps include grouping and factoring terms, completing the square for both x and y terms, and simplifying to reach the standard form: (x + 4)^2 / 36 - (y + 2)^2 / 9 = 1.

15:16

📘 Final Example and Conclusion

The final example discussed is 16x^2 + 9y^2 + 36y = 64x + 44. The video details the transformation process through completing the square and factoring, resulting in the equation (x - 2)^2 / 9 + (y + 2)^2 / 16 = 1. The video concludes by encouraging viewers to like, subscribe, and hit the bell button for more tutorials.

Mindmap

Keywords

💡Hyperbola

A hyperbola is a type of conic section that is formed by the intersection of a plane with a double cone. It is characterized by having two separate branches that are mirror images of each other. In the video, the hyperbola is the main focus, with the discussion centered on its standard form equation and how to manipulate it using algebraic techniques.

💡General Equation

The general equation refers to a form of an equation that includes all possible terms. In the context of the video, it is the starting point for the hyperbola's equation, which is given as 'Ax^2 + Bx + Cy^2 + Dy + E = 0'. The video aims to transform this general form into a standard form to better understand the properties of the hyperbola.

💡Standard Form

The standard form of an equation is a simplified version that makes it easier to identify and analyze the properties of the mathematical object it represents. For hyperbolas, the standard form is typically written as (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola. The video demonstrates how to convert the general equation of a hyperbola into this standard form.

💡Completing the Square

Completing the square is an algebraic technique used to transform a quadratic equation into a perfect square trinomial. This method is essential in the video for converting the general form of the hyperbola's equation into the standard form. It involves adding and subtracting the same value to make the x or y terms square-able.

💡Perfect Square Trinomial

A perfect square trinomial is a special form of quadratic equation that can be factored into the square of a binomial. In the video, the presenter uses this concept to simplify the terms of the hyperbola's equation, which is a crucial step in transforming the general equation into the standard form.

💡Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. They include circles, ellipses, parabolas, and hyperbolas. The video specifically discusses hyperbolas, which are one of the four types of conic sections, and how their equations can be manipulated to understand their geometric properties.

💡Algebraic Manipulation

Algebraic manipulation refers to the process of changing the form of an equation using various algebraic techniques without changing the meaning of the equation. In the video, the presenter demonstrates several algebraic manipulations, such as completing the square and dividing terms, to transform the general equation of a hyperbola into its standard form.

💡Dividing Terms

Dividing terms is an algebraic operation used to simplify equations by making the coefficients of the terms more manageable. In the context of the video, the presenter divides terms to normalize the equation of the hyperbola, which is necessary for it to fit the standard form.

💡Coefficients

In algebra, coefficients are the numerical factors in a term of an equation. They indicate the number of times a variable is multiplied by itself or added/subtracted. In the video, coefficients are used and manipulated to transform the general equation of the hyperbola into a more recognizable standard form.

💡Binomial

A binomial is a polynomial with two terms, such as 'x + y' or '3x - 5'. In the video, binomials are used in the context of completing the square, where the presenter forms binomials that can be squared to create a perfect square trinomial, which is then used to express the hyperbola's equation in standard form.

💡Equation Transformation

Equation transformation is the process of changing an equation from one form to another, often to make it easier to solve or analyze. The video is centered around transforming the general equation of a hyperbola into its standard form through a series of algebraic steps, including completing the square and dividing by a constant.

Highlights

Introduction to the general equation of a hyperbola in standard form.

Explanation of the general equation format: ax^2 + by^2 + cx + dy + e = 0.

First example: Transforming 4x^2 - y^2 - 24x = 64 into standard form.

Process of dividing by a factor to simplify the equation.

Completing the square for both x and y terms.

Resulting in the equation (x - 3)^2/25 - y^2/4 = 1.

Second example: Converting 3x^2 - y^2 + 18x + 4y = -35 into standard form.

Application of completing the square for the x term.

Final equation form: (x + 3)^2/4 + (y - 2)^2/12 = 1.

Third example: Transforming x^2 - 4y^2 + 8x - 16y - 36 = 0.

Completing the square for both x and y terms in the third example.

Resulting in the equation (x + 4)^2/36 - (y + 2)^2/9 = 1.

Fourth example: Working with 16x^2 + 9y^2 + 36y = 64x + 44.

Isolating and simplifying terms to prepare for completing the square.

Final equation form: (x - 2)^2/9 + (y + 2)^2/16 = 1.

Summary of the process and the importance of completing the square.

Encouragement to like, subscribe, and hit the bell for more video tutorials.

Transcripts

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