Pre-Calculus : Hyperbola - Transforming General Form to Standard Form
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
- 📚 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
📚 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.
🔍 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.
Mindmap
Keywords
💡Hyperbola
💡Standard Form
💡Center
💡General Form
💡Binomial
💡Coefficient
💡Regrouping
💡Perfect Square Trinomial
💡Simplification
💡Transformation
💡Equation Manipulation
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]
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