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

Summary

In 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

• 📚 The video is a tutorial on converting the general form of a hyperbola to its standard form.
• 📐 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 \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
• 📍 When the center is not at the origin, the standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
• 🔍 The given problem to solve is \( 4x^2 - 5y^2 + 32x + 30y = 1 \).
• ✂️ The first step is to regroup the terms in the equation to prepare for completing the square.
• 🔢 After regrouping, the equation becomes \( 4x^2 + 32x - 5y^2 - 30y = -13 \).
• 📈 Factor out the coefficients from the x and y terms to simplify the equation.
• 📊 Complete the square for both x and y terms by adding the necessary constants to maintain equality.
• 🔄 Adjust the equation by adding constants to both sides to form perfect square trinomials.
• 📉 The completed square form of the equation is \( (x + 4)^2 - 5(y - 3)^2 = 20 \).
• 🔄 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

00:00

📚 Introduction to Hyperbola Standard Form Transformation

The video begins with a welcome to Senior Pablo TV, where the tutorial focuses on converting the general form of a hyperbola to its standard form. The host provides a quick review of the standard forms of hyperbolas, including those centered at the origin and those with a center at (h, k). The problem presented involves transforming the equation 4x^2 - 5y^2 + 32x + 30y = 1 into standard form. The process starts by regrouping terms and factoring out common factors, aiming to create a perfect square trinomial for both x and y terms. The video demonstrates the algebraic steps required to achieve this, including dividing by the necessary constants to isolate the perfect square trinomials.

05:01

🔍 Completing the Square and Simplifying to Standard Form

In the second paragraph, the tutorial continues by completing the square for both x and y terms in the given hyperbola equation. The host divides the equation by 2 to find the middle term for the perfect square trinomials and then adds the necessary constants to both sides of the equation to maintain equality. After obtaining the perfect square trinomials, the equation is simplified by adding the constants to the right side and then dividing the entire equation by the sum of these constants to achieve the standard form. The final step is to simplify the coefficients to get the standard form of the hyperbola, which is presented with the equation (x + 4)^2/5 - (y - 3)^2/4 = 1. The video concludes with a summary of the steps and a thank you to the viewers.

Keywords

💡Hyperbola

A hyperbola is a type of conic section that resembles two mirror-imaged parabolas opening in opposite directions. In the context of the video, the hyperbola's standard form is discussed, which is essential for understanding how to transform a general equation into the standard form of a hyperbola. The script provides examples of the standard forms when the center is at the origin and when it is not.

💡Standard Form

The standard form of an equation is a specific arrangement that makes it easier to identify and work with the properties of geometric figures. For hyperbolas, the standard form helps in identifying the center, axes, and other characteristics. The video script explains how to convert a general equation into the standard form of a hyperbola, which is a key step in solving problems related to hyperbolas.

💡Center

The center of a hyperbola is the point (h, k) that equates to the origin (0, 0) when the hyperbola is centered there. The script discusses two scenarios: when the center is at the origin and when it is not, highlighting the different standard forms of hyperbolas in each case.

💡General Form

The general form of an equation is a more flexible representation that does not necessarily reveal the geometric properties of the figure immediately. In the script, the general form of a hyperbola is given, and the process of transforming it into the standard form is demonstrated, which is crucial for analyzing the hyperbola's characteristics.

💡Binomial

A binomial is an algebraic expression with two terms. In the context of the video, binomials are squared to form part of the hyperbola's equation. The script mentions 'a square of a binomial' when discussing the standard form of a hyperbola centered at the origin.

💡Coefficient

In algebra, coefficients are numerical factors that multiply variables in an equation. The script refers to coefficients when discussing the terms of the hyperbola's equation, such as '4x squared' and '-5y squared', which are crucial for the transformation process.

💡Regrouping

Regrouping in algebra involves rearranging terms in an equation to facilitate further manipulation. The script describes regrouping the terms of the given hyperbola equation to prepare for the transformation into the standard form.

💡Perfect Square Trinomial

A perfect square trinomial is a binomial squared, resulting in an equation of the form (ax + b)^2. The video script uses the concept of perfect square trinomials to transform the general form of the hyperbola into a form that resembles the standard form.

💡Simplification

Simplification in mathematics involves reducing an equation or expression to a simpler form without changing its value. The script demonstrates simplification steps, such as dividing the entire equation by a constant to achieve the standard form of the hyperbola.

💡Transformation

Transformation in the context of the video refers to the process of converting the general form of a hyperbola's equation into its standard form. This is a key concept as it allows for easier analysis and understanding of the hyperbola's properties.

💡Equation Manipulation

Equation manipulation involves algebraic techniques to modify an equation into a desired form. The script provides a step-by-step guide on how to manipulate the given hyperbola equation to reach its standard form, which is essential for solving related problems.

Highlights

Introduction to the tutorial on transforming general form to standard form of a hyperbola.

Recall of standard forms of hyperbolas with center at the origin.

Explanation of standard form variations for hyperbolas with different centers.

Presentation of the problem: transforming the equation 4x^2 - 5y^2 + 32x + 30y = 1 into standard form.

Step-by-step regrouping of the given equation to prepare for transformation.

Identification of the need for a common factor to simplify the equation.

Process of creating a perfect square trinomial for x and y components.

Division by two to find the middle term for the perfect square trinomial.

Addition of constants to both sides of the equation to complete the square.

Explanation of the necessity to balance the equation by adding constants.

Transformation of the equation into a form that resembles the standard hyperbola equation.

Division of the entire equation by a constant to achieve the standard form.

Final simplification of the equation to present the standard form of the hyperbola.

Conclusion of the tutorial with the final standard form of the hyperbola equation.

Acknowledgment of the audience and a sign-off from Senior Pablo TV.

Applause indicating the end of the tutorial video.

Transcripts

good day everyone welcome to senior

pablo tv

and for today's tutorial video we will

be discussing

transforming general form to standard

form of hyperbola

so let's have first a short recall about

the standard forms of the

hyperbola so if the center is zero zero

we have this form x squared over a

squared

minus y squared over b squared is equal

to one or it can be

y squared over a squared minus x squared

over b squared is equal to one

and e of the center is not located

at the origin we have

the standard form for hk

x our quantity x minus h raised to two

over a squared minus

quantity y minus k raised to two over b

squared is equal to one

so let's have this problem

4x squared minus 5y squared plus

32x plus 30y is equal to 1.

put in mind that in our

standard form in hk

and if the center is at

zero zero we have a square of a binomial

so

our first step is we need to

regroup the given

so we have 4 x squared

plus 32x

now we have

negative

5 y squared then

negative 3y is equal to 1

why it became negative 30

because negative times negative

is positive 13 and negative times

positive side negative 5

and now let's make it

let's get first the common factor

so 4

now we have x squared

plus 32 divided by 4

that is

[Music]

minus here is five

y squared minus

negative thirty

negative thirty

uh thirty divided by five

that is six six

y plus

blank is equal to one

now let us make

this a perfect square trinomial

so divide by two get the middle third

divided by two so eight divided by two

that is four square first square is

sixteen

so we need to add sixteen here

[Music]

and of course on the right side we need

two

so we have four here the four multiply

it

to the 16.

now let us make this a perfect square

trinomial

negative six divided by two that is

negative three negative three square

which is

positive nine of course

we need to add that into our right side

so cap is negative five

times nine

okay now we have

four

open parenthesis making the square of a

binomial

so x plus

square root of 16 for raised to

2 minus 5

y minus square root of 9

3 raised to 2 is equal to

1 plus 4 times 16

that is 64 minus

45

so simplify so copy

4 x plus 4 raised to 2

minus 5 y minus 3

raised to 2 is equal to

1 64 that is

65 minus 45

that is 20.

so in our standard form it must be equal

to one

so we need this to divide by

20 and by 20

so this is now equal to 1 so simplify

4 divided by 20

that is one over five or simply for our

answer

x plus four squared all over

five minus

five divided by twenty that is 1 4.

so our denominator is 4

y minus 3 squared

so this will be the standard form

of the hyperbola

so that's the step on how to transform

general form to a standard form of the

hyperbola thank you for watching senior

pablo tv

[Applause]