Problem 80 - Intermediate Algebra

Question

Write the equation \(x^{2}-4 y^{2}+2 x-8 y=7\) in standard form to show that it describes a hyperbola.

Step-by-Step Solution

Answer

The equation describes a hyperbola: \(\frac{(x+1)^2}{7} - \frac{(y+1)^2}{7/4} = 1\).

Step 1: Group Terms Appropriately

Rewrite the given equation by grouping the terms involving \(x\) together and \(y\) together.\[ x^2 + 2x - 4y^2 - 8y = 7 \]

Step 2: Complete the Square for x

For the \(x\) terms, factor out the coefficient of \(x^2\) if necessary (it's not necessary here, since it's 1) and complete the square.Take half of the coefficient of \(x\), which is \( \frac{2}{2} = 1 \), square it \((1)^2 = 1\), and add it inside the equation.\[ x^2 + 2x + 1 - 1 - 4y^2 - 8y = 7 \]

Step 3: Complete the Square for y

For the \(y\) terms, factor out \(-4\) from \(-4y^2 - 8y\) to get \(-4(y^2 + 2y)\) and complete the square inside the bracket. Take half of the coefficient of \(y\), which is \( \frac{2}{2} = 1 \), square it \((1)^2 = 1\), and add inside the square, making sure to multiply by the factor outside.\[- 4(y^2 + 2y + 1 - 1) = -4((y+1)^2 - 1) = -4(y+1)^2 + 4 \]

Step 4: Simplify and Rearrange

Substitute the completed squares back into the equation:\[ (x^2 + 2x + 1) - 1 - 4((y+1)^2) + 4 = 7 \]Simplify by combining constants:\[ (x+1)^2 - 4(y+1)^2 = 7 \]

Step 5: Write in Standard Form

Shift any unnecessary constants on the left side over to the right:\[ \frac{(x+1)^2}{7} - \frac{(y+1)^2}{7/4} = 1 \]This shows the standard form of a hyperbola.

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This method is not only useful in solving quadratic equations but also essential in expressing equations of conic sections, such as hyperbolas, in standard form. A quadratic expression typically looks like this:

• x quadratic form: \( x^2 + bx \)
• y quadratic form: \( y^2 + by \)

To complete the square for a term like \( x^2 + 2x \), you would:1. Take half of the linear coefficient, which is the coefficient of \( x \) or \( y \), divide it by 2.2. Square the result to obtain a perfect square trinomial.3. Add and subtract this squared number within the equation.
For example, taking the term \( x^2 + 2x \):- Half of 2 is 1. Squaring it gives \( 1^2 = 1 \).- Therefore, you add and subtract 1: \( (x + 1)^2 - 1 \).
This step-by-step process helps transform the quadratic into a form that is more understandable and usable for further algebraic manipulation.

Standard Form

When dealing with conic sections like hyperbolas, we aim to express the equation in a specific format, known as the standard form. The standard form of a hyperbola, centered at some point \((h, k)\), is given by:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]This format clearly shows:

• The center: Located at \( (h, k) \).
• The orientation: If the \( x \) part comes first, it indicates the transverse axis is horizontal.
• The lengths: \( a \) and \( b \) represent distances that influence the width and height of the hyperbola.

In the given exercise, rewriting the original equation into standard form involves algebraic manipulation and completing the square mechanism to achieve:\[ \frac{(x+1)^2}{7} - \frac{(y+1)^2}{7/4} = 1 \]By reorganizing into this form, we can easily identify key characteristics such as center and axes orientation.

Algebraic Manipulation

Algebraic manipulation is a crucial skill when working with equations, especially while converting them from general to standard form. This involves strategic operations such as addition, subtraction, multiplication, or division to simplify or rearrange expressions for results that are more meaningful or easier to work with.
In the context of the given problem, algebraic manipulation involves:

• Breaking down terms: Grouping \( x \) and \( y \) terms separately.
• Completing the square: As described, to turn binary quadratic forms into perfect squares.
• Shifting terms: Moving constants across the equal sign to balance the equation.
• Dividing through: Adjusting the equation so that it fits the standard conic section form, which involves dividing each completed square by the constant needed to make the right side equal 1.

Each step is executed cautiously to maintain balance and to ensure that the equations remain equivalent while gradually transforming them to reveal their geometric implications.