Problem 13 - Intermediate Algebra

Question

Divide both sides of the equation by 100 and write the equation in standard form: $$100(x+1)^{2}-25(y-5)^{2}=100$$

Step-by-Step Solution

Answer

\((x+1)^2 - \frac{1}{4}(y-5)^2 = 1\) is the simplified equation.

Step 1: Divide Both Sides by 100

We start with the equation: \(100(x+1)^{2} - 25(y-5)^{2} = 100\). We need to divide every term by 100 to simplify the equation. This gives us: \(\frac{100(x+1)^{2}}{100} - \frac{25(y-5)^{2}}{100} = \frac{100}{100}\). Simplifying each term, we have \((x+1)^{2} - \frac{1}{4}(y-5)^{2} = 1\).

Step 2: Rewrite Equation in Standard Form

The current form \((x+1)^2 - \frac{1}{4}(y-5)^2 = 1\) can be rewritten in standard form by ensuring each coefficient is clear and simplified. The equation represents a hyperbola in standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). From the equation, we identify \(h = -1\), \(k = 5\), \(a^2 = 1\), and \(b^2 = 4\). Thus, the standard form of the hyperbola is \(\frac{(x+1)^2}{1} - \frac{(y-5)^2}{4} = 1\).

Key Concepts

Standard Form

When dealing with equations representing conic sections like hyperbolas, putting them into standard form makes them easier to analyze and understand. The standard form of a hyperbola is given by:

• \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) for a hyperbola that opens left and right.
• \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \) for a hyperbola that opens up and down.

Here,

• • \(h\) and \(k\) are the coordinates of the center of the hyperbola.
  • \(a\) and \(b\) are related to the distances from the center to the vertices along the transverse and conjugate axes, respectively.

Placing the equation in standard form gives a clearer picture of these values, allowing us to identify the type of conic section and the orientation it exhibits.

Dividing Equations

Dividing each term of an equation by a constant is a crucial step in simplifying and reformatting expressions. In our exercise, dividing the entire equation by 100 helped us simplify it from

• \(100(x+1)^2 - 25(y-5)^2 = 100\)

to a more workable expression:

• \((x+1)^2 - \frac{1}{4}(y-5)^2 = 1\)

When you divide every part of an equation by the same number, you do not alter the equation's fundamental nature, as you maintain kinematic balance. This process helps shrink down coefficients, making patterns and variables clearer.
It is a tool used not just in hyperbolas but across many algebraic challenges.

Simplifying Equations

Simplification is key in solving any mathematical problem. In the context of our exercise, simplifying allowed us to move from a complicated expression to one that is ready for interpretation and analysis in its standard form.

• Initial Equation: \(\frac{100(x+1)^{2}}{100} - \frac{25(y-5)^{2}}{100} = \frac{100}{100}\)
• Simplified form: \((x+1)^{2} - \frac{1}{4}(y-5)^{2} = 1\)

This step of breaking down and reducing fractions or terms into their simplest forms helps focus on the core mathematical concepts without getting lost in numerical complexity. Remember:

• Combining like terms and reducing fractions are common techniques in simplification.
• This step also makes it easier to translate equations into their respective standard forms when analyzing conic sections like ellipses, hyperbolas, and parabolas.