Pre-Calculus : Hyperbola - Transforming General Form to Standard Form

SeΓ±or Pablo TV
30 Oct 202006:42

Summary

TLDRIn this tutorial from Senior Pablo TV, viewers are guided through the process of converting the general form of a hyperbola to its standard form. The video begins with a refresher on standard hyperbola forms, then tackles a specific problem involving algebraic manipulation to regroup terms and create perfect square trinomials. The final step involves adjusting the equation to match the standard form, showcasing the mathematical steps in a clear and concise manner. The tutorial concludes with the transformed equation and an appreciation for the viewers' attention.

Takeaways

  • πŸ“š The video is a tutorial on converting the general form of a hyperbola to its standard form.
  • πŸ“ The standard form of a hyperbola with the center at the origin is either \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
  • πŸ“ When the center is not at the origin, the standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
  • πŸ” The given problem to solve is \( 4x^2 - 5y^2 + 32x + 30y = 1 \).
  • βœ‚οΈ The first step is to regroup the terms in the equation to prepare for completing the square.
  • πŸ”’ After regrouping, the equation becomes \( 4x^2 + 32x - 5y^2 - 30y = -13 \).
  • πŸ“ˆ Factor out the coefficients from the x and y terms to simplify the equation.
  • πŸ“Š Complete the square for both x and y terms by adding the necessary constants to maintain equality.
  • πŸ”„ Adjust the equation by adding constants to both sides to form perfect square trinomials.
  • πŸ“‰ The completed square form of the equation is \( (x + 4)^2 - 5(y - 3)^2 = 20 \).
  • πŸ”„ To achieve the standard form, divide the entire equation by 20 to get \( \frac{(x + 4)^2}{5} - \frac{(y - 3)^2}{4} = 1 \).
  • 🎯 The final standard form of the hyperbola is \( \frac{(x + 4)^2}{5} - \frac{(y - 3)^2}{4} = 1 \), indicating the center, orientation, and shape of the hyperbola.

Q & A

  • What is the main topic of the video tutorial?

    -The main topic of the video tutorial is transforming the general form of a hyperbola to its standard form.

  • What are the standard forms of a hyperbola when the center is at the origin?

    -The standard forms of a hyperbola when the center is at the origin are \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

  • What is the standard form of a hyperbola when the center is not at the origin?

    -The standard form of a hyperbola when the center is not at the origin is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).

  • What is the given hyperbola equation in the problem?

    -The given hyperbola equation in the problem is \( 4x^2 - 5y^2 + 32x + 30y = 1 \).

  • What is the first step in transforming the given equation to the standard form?

    -The first step is to regroup the terms of the given equation to form a binomial square on both sides.

  • How does the script handle the term '32x' in the equation?

    -The script divides '32x' by the common factor '4' to simplify it to '8x'.

  • What is the purpose of adding '16' and '9' to the equation in the script?

    -The purpose of adding '16' and '9' is to complete the square for the 'x' and 'y' terms respectively, to transform the equation into a perfect square trinomial.

  • What is the final standard form of the hyperbola after the transformation?

    -The final standard form of the hyperbola is \( \frac{(x + 4)^2}{5} - \frac{(y - 3)^2}{4} = 1 \).

  • What is the significance of dividing the entire equation by '20' in the script?

    -Dividing the entire equation by '20' is done to normalize the equation so that the right side equals '1', which is a requirement for the standard form of a hyperbola.

  • What is the final step in the script to ensure the equation is in the standard form?

    -The final step is to simplify the coefficients of the squared terms and ensure the equation equals '1', resulting in the standard form of the hyperbola.

  • Why is it important to transform the general form of a hyperbola to its standard form?

    -Transforming the general form to the standard form is important because it allows for easier identification of the hyperbola's properties, such as its center, vertices, and asymptotes.

Outlines

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Mindmap

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Keywords

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Highlights

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Transcripts

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Hyperbola MathStandard FormGeneral FormMath TutorialEquation TransformationGeometry LessonEducational VideoAlgebraic GeometryMathematics EducationPablo TVMath Tips