Problem 19 - Thomas Calculus

Question

Exercises \(17-24\) give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch. $$ 2 x^{2}+y^{2}=2 $$

Step-by-Step Solution

Answer

The ellipse in standard form is \(\frac{x^2}{1} + \frac{y^2}{2} = 1\). The foci are \(\left(\pm \frac{1}{\sqrt{2}}, 0\right)\).

Step 1: Identify the given equation

The given equation of the ellipse is \(2x^2 + y^2 = 2\). This is the equation we need to convert into standard form.

Step 2: Write the standard form of ellipse equation

The standard form of an ellipse with a horizontal major axis is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), and with a vertical major axis is \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\). We need to rewrite the given equation to match one of these forms.

Step 3: Divide the entire equation by 2

To make the equation match the standard form where the right-hand side equals 1, divide every term in the equation \(2x^2 + y^2 = 2\) by 2, resulting in \(x^2 + \frac{y^2}{2} = 1\).

Step 4: Identify the values of semi-major and semi-minor axes

Comparing \(x^2 + \frac{y^2}{2} = 1\) with the standard form \(\frac{x^2}{1} + \frac{y^2}{1/2} = 1\), we determine that \(a^2 = 1\) and \(b^2 = \frac{1}{2}\). So, \(a = 1\) and \(b = \frac{1}{\sqrt{2}}\). The major axis is horizontal since \(a > b\).

Step 5: Find the coordinates of the foci

The foci of an ellipse are located at \((h \pm c, k)\) for a horizontal major axis, where \(c = \sqrt{a^2 - b^2}\). Here, \(c = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}\). Since \(h = 0\) and \(k = 0\), the foci are at \(\left(\pm \frac{1}{\sqrt{2}}, 0\right)\).

Step 6: Sketch the ellipse

Draw an ellipse centered at the origin (0,0) with a horizontal major axis. The semi-major axis is 1 unit long along the x-axis, and the semi-minor axis is \(\frac{1}{\sqrt{2}}\) along the y-axis. Mark the foci at \(\left(\pm \frac{1}{\sqrt{2}}, 0\right)\).

Key Concepts

Standard Form of an Ellipse

Ellipses have a standardized equation that helps us easily identify their properties. The standard form of an ellipse relies on whether its major axis is horizontal or vertical. For a horizontally oriented major axis, the equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). If the major axis is vertical, the equation becomes \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\). Here:

• \((h, k)\) is the center of the ellipse.
• \(a\) is the length of the semi-major axis.
• \(b\) is the length of the semi-minor axis.

To convert a given equation such as \(2x^2 + y^2 = 2\) into one of these standard forms, you need to simplify it so that the right side equals 1. For example, by dividing every term by the constant 2, you transform the equation to \(x^2 + \frac{y^2}{2} = 1\). This step allows you to identify the values of \(a\) and \(b\).

Foci of an Ellipse

The foci of an ellipse are two points along the major axis that help define its shape. They are pivotal in understanding the ellipse's unique properties, as they help determine how the curve stretches and bends. The distance from the center to a focus is denoted \(c\), calculated using the equation \(c = \sqrt{a^2 - b^2}\). For instance, with our earlier equation \(x^2 + \frac{y^2}{2} = 1\), we find that \(a = 1\) and \(b = \frac{1}{\sqrt{2}}\). Thus, \(c = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}\).The foci are located at \((h \pm c, k)\) for ellipses with horizontal major axes, or \((h, k \pm c)\) for those with vertical ones. In our scenario, centered at the origin \((0,0)\), and having a horizontal major axis, the foci are positioned at \(\left(\pm \frac{1}{\sqrt{2}}, 0\right)\). Understanding the foci is crucial for graphing and interpreting the ellipse's structure.

Sketching Ellipses

Sketching an ellipse involves plotting it accurately based on its standard form equation. You'll need to determine the orientation—whether horizontal or vertical—by comparing the squared lengths of the axes, \(a^2\) and \(b^2\). If \(a^2 > b^2\), the major axis is horizontal; otherwise, it's vertical.Begin by plotting the center \((h, k)\), and extend the semi-major axis \(a\) units in both directions along the major axis. Next, plot the ends of the semi-minor axis \(b\) units perpendicular to the major axis. For the equation \(x^2 + \frac{y^2}{2} = 1\), the center is at \((0,0)\), the semi-major axis length is 1 (on the x-axis), and the semi-minor is \(\frac{1}{\sqrt{2}}\) (on the y-axis).Finally, incorporate the foci into your sketch for completeness, confirming the ellipse's accurate shape. The foci \(\left(\pm \frac{1}{\sqrt{2}}, 0\right)\) are essential features in this visual representation.

Semi-Major and Semi-Minor Axes

An ellipse has two essential dimensions: the semi-major and semi-minor axes. These define its basic shape and proportions:

• The **semi-major axis** is the longest radius of the ellipse, running through the center to the perimeter.
• The **semi-minor axis** is the shorter radius, positioned perpendicular to the major axis.

For a given equation, values \(a\) and \(b\) in \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) determine the lengths of these axes. Specifically, \(a\) corresponds to the semi-major axis if it's the greater value compared to \(b\). For our equation \(x^2 + \frac{y^2}{2} = 1\), we have \(a = 1\) and \(b = \frac{1}{\sqrt{2}}\), so the major axis is horizontal.These axes are not just dimensions; they control how stretched or rounded the ellipse appears. Knowing the lengths of the semi-major and semi-minor axes offers a quick insight into the ellipse's geometric nature.