Problem 12 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 4 x^{2}+y^{2}=12 $$

Step-by-Step Solution

Answer

Vertices are \((0, \pm 2\sqrt{3})\), endpoints of minor axis \((\pm \sqrt{3}, 0)\), and foci \((0, \pm 3)\).

Step 1: Write the Equation in Standard Form

The given equation of the ellipse is \(4x^2 + y^2 = 12\). The equation of an ellipse in standard form is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). First, divide the entire equation by 12 to arrange it into the standard form.\(\frac{4x^2}{12} + \frac{y^2}{12} = 1\) reduces to \(\frac{x^2}{3} + \frac{y^2}{12} = 1\). This is now in standard form with \(a^2 = 12\) and \(b^2 = 3\).

Step 2: Identify the Vertices

In the equation \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\), the vertices are located at \((h, k \pm a)\). Here we can see that \(a^2 = 12\), hence \(a = \sqrt{12} = 2\sqrt{3}\). Thus, the vertices are \((0, \pm 2\sqrt{3})\).

Step 3: Identify the Endpoints of the Minor Axis

The endpoints of the minor axis are located at \((h \pm b, k)\). Here \(b^2 = 3\), so \(b = \sqrt{3}\). Therefore, the endpoints of the minor axis are \((\pm \sqrt{3}, 0)\).

Step 4: Find the Foci

The foci of an ellipse are found using the relationship \(c^2 = a^2 - b^2\). Here, \(a^2 = 12\) and \(b^2 = 3\), so \(c^2 = 12 - 3 = 9\) which gives \(c = 3\). Therefore, the foci are \((0, \pm 3)\).

Step 5: Sketch the Graph

Use the calculated vertices \((0, \pm 2\sqrt{3})\), endpoints of the minor axis \((\pm \sqrt{3}, 0)\), and foci \((0, \pm 3)\) to sketch the graph of the ellipse on a coordinate plane. Draw an ellipse centered at the origin that stretches vertically more than it does horizontally.

Key Concepts

Standard Form of an Ellipse

The standard form of an ellipse equation helps us to easily identify its major and minor axes, as well as other properties. The formula is given by \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where

• \((h, k)\) is the center of the ellipse.
• \(a^2\) represents the square of the semi-major axis length if \(a > b\). Conversely, it is the square of the semi-minor axis if \(b > a\).
• \(b^2\) is the square of the semi-minor axis length if \(a > b\), or the square of the semi-major axis if \(b > a\).

To convert any given ellipse equation to this form, you divide all terms by the constant on the right side to make it equal 1. This step is crucial because it enables us to determine all the key components of the ellipse accurately. In the example, this is achieved by dividing the equation \(4x^2 + y^2 = 12\) by 12, resulting in \(\frac{x^2}{3} + \frac{y^2}{12} = 1\). Here, we can interpret that our major axis is along the y-axis and our ellipse is vertically oriented.

Vertices of an Ellipse

Vertices of an ellipse are the points that lie on the longest diameter of the ellipse, known as the major axis. For a standard form equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), if \(a^2 > b^2\), the vertices are located at \((h, k\pm a)\). If \(b^2 > a^2\), then they are at \((h\pm b, k)\).
In the given example, our equation \(\frac{x^2}{3} + \frac{y^2}{12} = 1\) identifies \(a^2 = 12\) and \(b^2 = 3\), meaning the major axis is vertical. Thus, the vertices are found by calculating \(a\), so, \(a = \sqrt{12} = 2\sqrt{3}\). The vertices' coordinates then become \((0, \pm 2\sqrt{3})\), indicating the maximum extent of the ellipse along the y-axis.

Minor Axis of an Ellipse

The minor axis of an ellipse is the shortest diameter that runs perpendicular to the major axis. For standard form equations such as \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] if \(a^2 > b^2\), the endpoints of the minor axis are given by \((h \pm b, k)\). Conversely, for \(b^2 > a^2\), they are \((h, k \pm a)\).
In our case, since \(b^2 = 3\), \(b = \sqrt{3}\). This turns the minor axis endpoints into \((\pm \sqrt{3}, 0)\). These points represent the shortest extent across the ellipse along the x-axis, showing the lateral size of the ellipse.

Foci of an Ellipse

The foci are two special points inside the ellipse. The sum of the distances from any point on the ellipse to each focus is constant. To determine the foci of an ellipse, we need to find \(c\), where \[ c^2 = a^2 - b^2 \] Using the values from the standard form of the ellipse, calculate \(c\) with \(a^2 = 12\) and \(b^2 = 3\). This results in \[ c^2 = 12 - 3 = 9 \] Thus, \(c = 3\).
Given that the ellipse's major axis is vertical, the foci are located along this axis at the points \((0, \pm 3)\). The foci provide insight into the ellipse's eccentricity, showing how stretched the ellipse is along its major axis.