Problem 18 - Precalculus: Mathematical for Calculus

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

The ellipse has vertices at \((0, \pm\sqrt{5})\), foci at \((0, \pm 2)\), and eccentricity \(e = \sqrt{\frac{4}{5}}\).

Step 1: Rearrange the Equation

Start by rearranging the given ellipse equation in standard form. The given equation is \(20x^2 + 4y^2 = 5\). Divide every term by 5 to get \(\frac{20x^2}{5} + \frac{4y^2}{5} = 1\), which simplifies to \(4x^2 + \frac{4}{5}y^2 = 1\).

Step 2: Simplify to Standard Ellipse Form

Divide the entire equation \(4x^2 + \frac{4}{5}y^2 = 1\) by 4 to get \(x^2 + \frac{1}{5}y^2 = \frac{1}{4}\). Then multiply through by 4 to further simplify: \(4x^2 + \frac{4}{5}y^2 = 1\) is now equivalent to \(\frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{5}{4}} = 1\), making it easier to identify the axes lengths.

Step 3: Identify Major and Minor Axes

In the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 < b^2\), the major axis is along the y-axis. Here, \(a^2 = \frac{1}{4}\) and \(b^2 = \frac{5}{4}\). Therefore, the major axis length is \(2b = 2\times\sqrt{\frac{5}{4}} = \sqrt{5}\), and the minor axis length is \(2a = 2\times\sqrt{\frac{1}{4}} = 1\).

Step 4: Find the Eccentricity

The eccentricity \( e \) of an ellipse is calculated using \( e = \sqrt{1 - \frac{a^2}{b^2}} \). Substituting the values \( a^2 = \frac{1}{4} \) and \( b^2 = \frac{5}{4} \), we find \( e = \sqrt{1 - \frac{1}{4}/\frac{5}{4}} = \sqrt{\frac{4}{5}} \).

Step 5: Locate the Foci of the Ellipse

For a vertical ellipse, the foci are located at \((0, \pm c)\), where \(c = be = \sqrt{5}/ \cdot \sqrt{\frac{4}{5}} = \sqrt{4} = 2 \). Thus, the foci are at \((0, 2)\) and \((0, -2)\).

Step 6: Identify the Vertices

The vertices of a vertical ellipse are at \((0, \pm b)\) where \(b = \sqrt{\frac{5}{4}}\). Therefore, the vertices are at \((0, \sqrt{5})\) and \((0, -\sqrt{5})\).

Step 7: Sketch the Ellipse

The ellipse is centered at the origin. Sketch the ellipse with the major axis along the y-axis, with vertices at \((0, \sqrt{5})\), \((0, -\sqrt{5})\), and minor axis parallel to the x-axis.

Key Concepts

Vertices of Ellipse

In geometry, the vertices of an ellipse are the points where the ellipse intersects its major axis. These points are significant as they represent the longest diameter of the ellipse. The equation of the ellipse we are examining is already arranged in its standard form: \[ \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{5}{4}} = 1 \]. For a vertical ellipse like this one, the major axis is aligned along the y-axis.

• The vertices occur at \((0, \pm b)\), where \( b = \sqrt{\frac{5}{4}} = \sqrt{5}/2 \).
• This means the vertices are located at \((0, \sqrt{5}/2)\) and \((0, -\sqrt{5}/2)\).

Understanding the position of the vertices helps to draw the outermost points of the ellipse, thereby aiding in sketching its correct shape.

Foci of Ellipse

The foci are crucial in defining the shape and properties of an ellipse. For any point on the ellipse, the total distance to the two foci is constant. The foci provide the ellipse its elliptical shape, differentiating it from circles. In our context, with a vertical ellipse, the foci are located along the major axis.To find the foci, we calculate the value \( c \), where \( c = b \cdot e \) and \( e \) is the eccentricity.

• Given \( b = \sqrt{5}/2 \) and \( e = \sqrt{\frac{4}{5}} \), \( c = \frac{\sqrt{5} \cdot \sqrt{4}}{2} = 2 \).
• Hence, the coordinates of the foci are \((0, 2)\) and \((0, -2)\).

These points play an important role in many applications such as in optics, where they relate to the path light takes when reflecting inside an elliptical mirror.

Eccentricity of Ellipse

The eccentricity is a measure of how "stretched" the ellipse is. It is a unitless number between 0 and 1, where 0 corresponds to a circle, and values closer to 1 signify a more elongated shape. For our ellipse, the eccentricity \( e \) is calculated using the formula:\[ e = \sqrt{1 - \frac{a^2}{b^2}} \] where \( a = \sqrt{\frac{1}{4}} = \frac{1}{2} \) and \( b = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \).

• Substituting these values, \( e = \sqrt{1 - \frac{\left(\frac{1}{2}\right)^2}{\left(\frac{\sqrt{5}}{2}\right)^2}} = \sqrt{\frac{4}{5}} \).
• This indicates the ellipse is moderately elongated, instead of circular.

Knowing the eccentricity helps us understand the shape of the ellipse, which is important in fields such as astronomy, where celestial orbits are elliptical.

Major Axis

The major axis is one of the defining features of an ellipse. It is the longest diameter that runs through the center of the ellipse, along which the vertices and foci are symmetrically placed. In a vertical ellipse, the major axis is aligned along the y-axis.

• For the given equation \( \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{5}{4}} = 1 \), the length of the major axis is calculated as \( 2b \), where \( b = \sqrt{\frac{5}{4}} \).
• Thus, the major axis length is \( 2 \times \frac{\sqrt{5}}{2} = \sqrt{5} \).

By knowing the major axis, we identify the longest dimension of the ellipse, providing a clear picture of its symmetry and geometric properties.

Minor Axis

The minor axis of an ellipse is the shortest diameter and is perpendicular to the major axis. This diameter touches the ellipse at its narrowest part, parallel to the x-axis in a vertical ellipse.

• Within our standard form equation, \( \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{5}{4}} = 1 \), the length of the minor axis is \( 2a \), where \( a = \sqrt{\frac{1}{4}} \).
• This provides us a minor axis length of \( 2 \times \frac{1}{2} = 1 \).

Having the measurements for both the major and minor axes is essential for accurately sketching an ellipse and understanding its layout relative to its surrounding plane.