Problem 34 - College Algebra

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. $$ 4 x^{2}+9 y^{2}=1 $$

Step-by-Step Solution

Answer

Center at (0,0); vertices at (-1/2, 0), (1/2, 0); foci at (-√5/6, 0), (√5/6, 0).

Step 1: Identify the Standard Form

The equation of the ellipse is given as \( 4x^{2} + 9y^{2} = 1 \). To identify the standard form, we rewrite the equation as \( \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1 \), which aligns with the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) with the center at \((h, k)\).

Step 2: Identify Center of the Ellipse

From the standard form \( \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1 \), we can see that \( h = 0 \) and \( k = 0 \) because there are no \( h \) or \( k \) values to shift the ellipse. Thus, the center of the ellipse is at the origin \((0,0)\).

Step 3: Determine the Lengths of the Semi-Axes

The semi-major axis length \( a \) is defined by the larger denominator. Here, \( b^2 = \frac{1}{9} \) so \( b = \frac{1}{3} \) and \( a^2 = \frac{1}{4} \) so \( a = \frac{1}{2} \). Thus, the lengths are \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \).

Step 4: Determine the Vertices

The vertices are located \( a \) units away from the center along the x-axis, and \( b \) units along the y-axis. Therefore, the vertices are at \((-\frac{1}{2}, 0), (\frac{1}{2}, 0)\) (along the major axis, the x-axis) and \((0, -\frac{1}{3}), (0, \frac{1}{3})\) (along the minor axis, the y-axis).

Step 5: Find the Foci

The foci distance from the center is determined using \( c^2 = a^2 - b^2 \). Substitute \( a^2 = \frac{1}{4} \) and \( b^2 = \frac{1}{9} \): \( c^2 = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \). Thus, \( c = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \). The foci are located at \((-\frac{\sqrt{5}}{6}, 0)\) and \((\frac{\sqrt{5}}{6}, 0)\) along the major axis.

Step 6: Plot the Ellipse

Draw the ellipse centered at the origin \((0,0)\). Mark the vertices at \((-\frac{1}{2}, 0), (\frac{1}{2}, 0)\), the co-vertices at \((0, -\frac{1}{3}), (0, \frac{1}{3})\), and the foci at \((-\frac{\sqrt{5}}{6}, 0)\) and \((\frac{\sqrt{5}}{6}, 0)\) on the graph.

Key Concepts

Standard Form of an Ellipse

Understanding the standard form of an ellipse is crucial when graphing it. An ellipse's standard equation looks like this:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]The general equation shifts the ellipse from its origin based on values \(h\) and \(k\). Here, \(a^2\) and \(b^2\) correspond to the semi-major and semi-minor axes, respectively. To create an ellipse, ensure the sum equals 1.

The exercise started with the equation \(4x^{2} + 9y^{2} = 1\), which needs adjustment to fit the standard form. We can rewrite it as:\[ \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1 \]Now, the equation matches the standard form with \(a^2 = \frac{1}{4}\) and \(b^2 = \frac{1}{9}\). This structure tells us the ellipse is aligned with both axes and has a center at the origin.

Center of an Ellipse

Determining the center of an ellipse is the foundation for graphing and understanding its symmetrical properties. The center, denoted as \((h, k)\), is identified from the general standard form equation:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]In our example, since there are no \((x-h)\) or \((y-k)\) components (it's just \(x^2\) and \(y^2\) in the equation), it means \(h = 0\) and \(k = 0\). Therefore, the ellipse is centered at the origin \((0, 0)\).

Understanding the center's coordinates helps locate vertices and foci easily. Whenever an ellipse appears to have a simple \(x^2\) and \(y^2\), like in our example, you can directly ascertain it is centered at the origin.

Vertices and Foci of Ellipses

Vertices and foci are critical points that define the shape and orientation of an ellipse. Let's break them down in simple terms:

- **Vertices:** These are the farthest and closest points along the major axis of the ellipse from the center. The major axis is where the largest denominator (from \(a^2\) or \(b^2\)) is located in your equation. - In our equation \(\frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1\), \(a^2 = \frac{1}{4}\) (so major axis on x-axis) and \(b^2 = \frac{1}{9}\). - Therefore, vertices are \((-\frac{1}{2}, 0)\) and \((\frac{1/2}, 0)\) for the major axis (x-axis), and \((0, -\frac{1}{3})\) and \((0, \frac{1}{3})\) for the minor axis (y-axis).- **Foci:** These are two special points on the major axis inside the ellipse. They are calculated with the formula: \[ c = \sqrt{a^2 - b^2} \] - In our case, substituting values gives \(c = \sqrt{\frac{1}{4} - \frac{1}{9}} = \frac{\sqrt{5}}{6}\). - This shows the foci are located at \((-\frac{\sqrt{5}}{6}, 0)\) and \((\frac{\sqrt{5}}{6}, 0)\) along the x-axis.Grasping these concepts simplifies sketching the ellipse and aids in understanding its spatial properties and symmetry.