Problem 29 - Precalculus Mathematics for Calculus

Question

Graph the equation \(4 x^{2}+2 y^{2}=1\) by solving for \(y\) and graphing two equations corresponding to the negative and positive square roots. (This graph is called an ellipse.)

Step-by-Step Solution

Answer

Graph \( y = \pm\sqrt{\frac{1}{2} - 2x^2} \) to form an ellipse.

Step 1: Rewrite the equation

The given equation is \( 4x^2 + 2y^2 = 1 \). To solve for \( y \), we first want to isolate \( y^2 \). Divide every term in the equation by 2 to simplify it: \( 2x^2 + y^2 = \frac{1}{2} \).

Step 2: Solve for y²

Rearrange the simplified equation from Step 1 to solve for \( y^2 \). Start by subtracting \( 2x^2 \) from both sides: \( y^2 = \frac{1}{2} - 2x^2 \).

Step 3: Take the square root

To solve for \( y \), take the square root of both sides of the equation. Remember to consider both the positive and negative roots:\( y = \pm \sqrt{\frac{1}{2} - 2x^2} \).

Step 4: Graph each equation

Now graph the two equations:- For the positive square root: \( y = \sqrt{\frac{1}{2} - 2x^2} \).- For the negative square root: \( y = -\sqrt{\frac{1}{2} - 2x^2} \).These will represent the top and bottom halves of the ellipse respectively.

Step 5: Identify the characteristics of the ellipse

Notice the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is an ellipse. Rewriting our equation in this form, we see:\( \frac{x^2}{\frac{1}{8}} + \frac{y^2}{\frac{1}{2}} = 1 \), identifying \( a = \frac{1}{\sqrt{8}} \) and \( b = \frac{1}{\sqrt{2}} \).

Key Concepts

Graphing Equations

Graphing equations is a fundamental concept in algebra and geometry. It translates mathematical relationships into visual representations. When you graph an equation, you're drawing a picture of all the possible solutions that satisfy it.
For example, graphing an ellipse involves plotting points that fit the equation's specific shape and pattern. In the given exercise, we have the equation \[4x^2 + 2y^2 = 1\]. Here, by focusing on graphing, we create a visual of this equation using its geometric form.
This geometric form helps us see how variables relate. In this instance, it helps show how changes in \(x\) and \(y\) relate to forming an ellipse. The ellipse will appear as a smooth, curved shape on the plane, with different parts of the equation affecting its width and height. Use graphing to help unravel these relationships visually.

Solving for y

Solving for a variable means isolating it on one side of an equation to find its relationship with other variables.
For the equation \(4x^2 + 2y^2 = 1\), we aim to solve for \(y\). The goal is to isolate \(y^2\) first, which requires simplifying the equation appropriately.
Start by dividing the entire equation by \(2\): \[2x^2 + y^2 = \frac{1}{2}\]. Now, subtract \(2x^2\) from both sides to further isolate \(y^2\): \[y^2 = \frac{1}{2} - 2x^2\].
This rearrangement is crucial as it sets the stage for understanding \(y\)'s effect on this elliptical pattern. It then allows us to solve for \(y\) itself by taking the square root of both sides. Remember always to consider both positive and negative roots because they are each part of the complete solution.
This dual solution is important as it maps onto the ellipse's top half (positive root) and bottom half (negative root) respectively.

Square Roots

Taking square roots is essential when solving equations that include squared terms, as they help undo the squaring and find the original value. In the exercise, once we have \(y^2 = \frac{1}{2} - 2x^2\), we take the square root on both sides to solve for \(y\).
This results in two outputs:

• Positive square root: \(y = \sqrt{\frac{1}{2} - 2x^2}\)
• Negative square root: \(y = -\sqrt{\frac{1}{2} - 2x^2}\)

These two equations represent the dual nature of square roots, forming two distinct parts of the ellipse.
The positive root gives the top section above the x-axis, while the negative covers the section below. Square roots are pivotal here for accurately capturing the complete geometrical figure and ensuring we fully understand its extent on the graph.
Always remember both signs are crucial; leaving one out would mean missing half the solutions.