Problem 18 - Algebra and Trigonometry

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$ 20 x^{2}+4 y^{2}=5 $$

Step-by-Step Solution

Answer

Vertices: (0, ±√1.25); Foci: (0, ±1); Eccentricity: ≈0.89; Major axis length: ≈2.24; Minor axis length: 1.

Step 1: Write the standard form of the ellipse equation

The given equation is \( 20x^{2} + 4y^{2} = 5 \). To convert it into the standard form, divide the entire equation by 5 to get \( \ \frac{20x^{2}}{5} + \frac{4y^{2}}{5} = 1 \. \) Simplifying gives \( \ \frac{4x^{2}}{1} + \frac{y^{2}}{1.25} = 1 \. \) Rearranging yields \( \ \frac{x^{2}}{0.25} + \frac{y^{2}}{1.25} = 1 \. \)

Step 2: Identify the semi-major and semi-minor axes

The standard form of an ellipse is \( \ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \ .\) Based on the equation \( \ \frac{x^{2}}{0.25} + \frac{y^{2}}{1.25} = 1 \ , \) we identify \(a^{2} = 1.25\) and \(b^{2} = 0.25\). Thus, \(a = \sqrt{1.25}\) and \(b = \sqrt{0.25}\).

Step 3: Calculate the vertices and foci

The vertices correspond to points along the major and minor axes. Here, the major axis is the y-axis (since 1.25 > 0.25). Therefore, vertices are \((0, \, \pm\sqrt{1.25}) \). The foci can be calculated using \(c = \sqrt{a^{2} - b^{2}} = \sqrt{1.25 - 0.25}\), yielding \(c = 1\). Hence, the foci are \((0, \, \pm 1)\).

Step 4: Determine the eccentricity and axes lengths

The eccentricity \( e \) of the ellipse can be calculated using \( e = \frac{c}{a} \), which gives \( e = \frac{1}{\sqrt{1.25}} \approx 0.89 \). The length of the major axis is \( 2a = 2\sqrt{1.25} \approx 2.24 \), and the minor axis is \( 2b = 2\sqrt{0.25} = 1 \).

Step 5: Plot the ellipse

To sketch the graph, plot the center at the origin, vertices at \((0, \, \pm\sqrt{1.25})\), and foci at \((0, \, \pm 1)\). The ellipse should extend further along the y-axis, as it's the major axis. Draw a symmetric oval shape centered at the origin.

Key Concepts

Vertices of an Ellipse

In the context of ellipses, vertices are the points where the ellipse is widest along its axes. These special points signify the farthest distance from the center along the principal axis or axes.

To find vertices, an ellipse needs to be expressed in its standard form equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively.

The vertices lie along the axes of greatest and least curvature, which means they are located at \((0, \pm a)\) or \((\pm a, 0)\), depending on whether the major axis is vertical or horizontal. In our example equation, the major axis is vertical because \(b^2 > a^2\). Thus, the vertices are at \((0, \pm \sqrt{1.25})\).

Understanding the positions of these vertices is crucial as it provides a visual reference for sketching the ellipse accurately, highlighting how it stretches along certain axes.

Foci of an Ellipse

The foci (singular: focus) are two essential points inside every ellipse. Their unique property is that the sum of the distances from any point on the ellipse to both foci is constant. This characteristic is intrinsic to the ellipse's definition.

To locate the foci, we use the relationship \( c = \sqrt{a^2 - b^2} \), which gives us the distance from the center to each focus. In ellipses, \(a\) represents the semi-major axis length, whereas \(b\) is the semi-minor axis length.

For our exercise, with a vertical major axis, the coordinates of the foci are given as \((0, \pm c)\). Substituting the values, we find \(c = \sqrt{1.25 - 0.25} = 1\), making the foci \((0, \pm 1)\).

Grasping the location of the foci helps in better understanding how the ellipse is structured and illustrates the balance within its geometric nature.

Eccentricity of an Ellipse

Eccentricity in the context of ellipses is a measure of how "elongated" the shape is compared to a circle. While a perfect circle has an eccentricity of 0, all ellipses have eccentricities between 0 and 1.

Eccentricity is calculated as \( e = \frac{c}{a} \), where \(c\) is the distance from the center to a focus, and \(a\) represents the length of the semi-major axis. The closer \(e\) is to 0, the more the ellipse resembles a circle.

In our retained example, the computed eccentricity is \( e = \frac{1}{\sqrt{1.25}} \approx 0.89 \). Such a value infers that the ellipse is quite elongated, extending more along its major axis.

Understanding the eccentricity is valuable because it quantitatively describes the ellipse's shape, thus helping in distinguishing between nearly circular and more stretched ellipses.