Problem 65 - Algebra for College Students

Question

For the indicated variable. Assume that all letters represent positive numbers. \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) for \(x\)

Step-by-Step Solution

Answer

\(x = \pm \sqrt{a^2 - \frac{a^2 y^2}{b^2}}\)

Step 1: Isolate the x-term

The given equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). First, we need to isolate the term containing \(x^2\). Subtract \(\frac{y^2}{b^2}\) from both sides: \[\frac{x^2}{a^2} = 1 - \frac{y^2}{b^2}.\]

Step 2: Clear the Denominator

Multiply both sides of the equation by \(a^2\) to eliminate the fraction: \[ x^2 = a^2 \left(1 - \frac{y^2}{b^2}\right). \]

Step 3: Simplify the Expression

Distribute the \(a^2\) on the right side: \[ x^2 = a^2 - \frac{a^2 y^2}{b^2}. \]

Step 4: Solve for x

Take the square root of both sides to solve for \(x\): \[ x = \pm \sqrt{a^2 - \frac{a^2 y^2}{b^2}}. \] Simplify further, if possible, recognizing that we assume all variables are positive.

Key Concepts

Isolating Variables

Isolating variables is a fundamental step in solving equations where we want to find the value of a specific variable. In our exercise, we began with the equation of an ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] The goal was to solve for \(x\), which means we needed to isolate the \(x^2\) term.

• Starting by subtracting \( \frac{y^2}{b^2} \) from both sides: this helps us get all terms involving \(y\) on one side and the \(x\) terms on the other.
• The equation becomes \( \frac{x^2}{a^2} = 1 - \frac{y^2}{b^2} \).

By isolating \(x^2\), we can then work towards directly solving for \(x\), making the complex equation easier to handle. The process of isolating a variable often involves algebraic operations such as addition, subtraction, and multiplication, and it is a critical first step in making sense of the equation.

Solving for a Variable

Once the variable is isolated, the next step is to solve for that variable, \(x\) in this case. After isolating the \(x^2\) term, our equation looked like this:\[ \frac{x^2}{a^2} = 1 - \frac{y^2}{b^2}. \] Now, to solve for \(x\), we had to clear the fractions first by multiplying both sides by \(a^2\) to eliminate the denominator, resulting in:\[ x^2 = a^2 \left(1 - \frac{y^2}{b^2}\right). \]Next, we distributed \(a^2\) over the terms within the parentheses, giving us:\[ x^2 = a^2 - \frac{a^2 y^2}{b^2}. \]To find \(x\), we take the square root of both sides of the equation. Importantly, remember when taking a square root, the solution can be both positive and negative so we have:\[ x = \pm \sqrt{a^2 - \frac{a^2 y^2}{b^2}}. \]As we assumed all variables are positive in the exercise, this can often allow us to simplify whether one or both of the solutions are applicable.

Conic Sections

Conic sections are fundamental in understanding various geometric shapes, and an ellipse is one of them. The general form of an ellipse's equation is:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] This describes an ellipse which is symmetric about both the \(x\)-axis and \(y\)-axis. It tells you how the ellipse stretches along these axes:

• The value \(a\) is the semi-major axis if \(a > b\), or the semi-minor axis if the reverse is true.
• Similarly, \(b\) represents the semi-major or semi-minor axis depending on their relationship with \(a\).

Ellipses have unique properties based on their equations and understanding these shapes' mathematical foundation helps in a variety of fields, from physics to engineering. Solving problems involving ellipses frequently requires skills in algebra, such as isolating and solving for variables, and a solid understanding of conic sections. This forms a bridge between abstract algebraic techniques and practical geometrical applications.