Problem 79 - Intermediate Algebra

Question

Write the equation \(x^{2}-y^{2}-2 x+4 y=12\) in standard form to show that it describes a hyperbola.

Step-by-Step Solution

Answer

The equation \((x-1)^2 - (y-2)^2 = 9\) in standard form is a hyperbola.

Step 1: Rewrite the Equation

Start with the given equation: \[ x^2 - y^2 - 2x + 4y = 12 \] We want to rearrange it into the standard form of a hyperbola, which is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). First, we'll complete the square for the \(x\) terms and the \(y\) terms separately.

Step 2: Complete the Square for x

For the \(x\) terms, focus on \(x^2 - 2x\). Complete the square:Rewrite it as:\[ x^2 - 2x = (x^2 - 2x + 1) - 1 = (x-1)^2 - 1 \] This process added and subtracted 1 to complete the square.

Step 3: Complete the Square for y

For the \(y\) terms, focus on \(-y^2 + 4y\). First, factor out \(-1\) to make it easier:Rewrite it as:\[ -y^2 + 4y = -(y^2 - 4y) \]Now complete the square inside the parentheses:\[ y^2 - 4y = (y^2 - 4y + 4) - 4 = (y-2)^2 - 4 \]In terms of \(\- (y-2)^2 + 4\), factoring the \(-1\) way.

Step 4: Substitute Completed Squares into the Original Equation

Substitute the completed squares back into the original equation:\[ (x-1)^2 - 1 - ((y-2)^2 - 4) = 12 \]Simplify further by adjusting the equation:\[ (x-1)^2 - 1 - (y-2)^2 + 4 = 12 \]Simplify the constants:\[ (x-1)^2 - (y-2)^2 + 3 = 12 \]Then, bring the constant on the right:\[ (x-1)^2 - (y-2)^2 = 12 - 3 \]\[ (x-1)^2 - (y-2)^2 = 9 \]

Step 5: Write in Standard Form of a Hyperbola

The equation \[(x-1)^2 - (y-2)^2 = 9\] is almost in standard form. Divide everything by 9 to normalize it:\[ \frac{(x-1)^2}{9} - \frac{(y-2)^2}{9} = 1 \]Now, the equation is in the standard form of a hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) with \(h = 1\), \(k = 2\), \(a^2 = 9\), and \(b^2 = 9\).

Key Concepts

Completing the Square

Completing the square is a method often used in algebra to make quadratic expressions more manageable, especially when dealing with conic sections like hyperbolas. It's essentially about transforming a part of the equation into a perfect square trinomial, which makes it easier to factor or rearrange. For example, given an expression like \[ x^2 - 2x \], we want to add and subtract the same value to make it a perfect square. The goal here is to have something of the form \[(x-a)^2\].

• Identify the coefficient of \(x\), which is -2 in this case.
• Halve this coefficient giving -1, then square it, resulting in 1.
• Add and subtract this squared result inside the equation. It looks like: \( (x^2 - 2x + 1) - 1 \).
• Factor the trinomial as: \((x-1)^2\), leaving the equation as: \((x-1)^2 - 1\).

This technique simplifies expressions significantly and is particularly useful when rewriting quadratic equations into their standard forms.
Similarly, when you have \(-y^2 + 4y\), it's best to first factor out the negative sign before completing the square. Once done, you perform similar steps to achieve a perfect square trinomial.

Standard Form of a Hyperbola

The standard form of a hyperbola's equation is crucial for understanding its properties. The typical formula is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \].Here is what each component represents:

• \((h, k)\) represents the center of the hyperbola.
• \(a\) and \(b\) are real numbers that indicate the distances that help shape the hyperbola.
• \(a^2\) and \(b^2\) are the denominators that dictate the spread of the hyperbola on the x-axis and y-axis respectively.
• The minus sign indicates the orientation and general shape of the hyperbola, confirming that it opens left and right along the x-axis.

Transforming the equation into this standard form helps in identifying the hyperbola's characteristics. It's about isolating the terms before and after completing the square, and adjusting constants around the equation to match the 1 on the right side.
By dividing the entire equation by 9 as shown in the solution, the hyperbola gets correctly standardized, paving the way for interpretation of its orientation and center.

Coordinate Geometry

Coordinate geometry is the mathematical study that explores geometric problems using the coordinate plane. This topic integrates algebraic methods and geometric principles to solve problems and define shapes like hyperbolas precisely.

• Key components involve coordinates of points, lines, and conic sections; specifically the equation for each.
• A hyperbola is a type of conic section, which can be visualized when examined in the xy-plane.
• Recognizing a hyperbola's orientation, center, and axis of symmetry is much easier once the equation is transformed into its standard form.
• In our exercise, recognizing the center of \((h, k) = (1, 2)\) is derived directly from matching the standard form to portions of the equation.

Coordinate geometry also aids in understanding the relationship between different segments of the hyperbola, such as vertices, transverse axes, and conjugate axes. Overall, it applies algebraic manipulation alongside geometric intuition to provide a powerful tool for analyzing the structure and properties of hyperbolas and other geometric entities.