Problem 13 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 4 x^{2}+\left(y+\frac{1}{2}\right)^{2}=4 $$

Step-by-Step Solution

Answer

The center is (0, -1/2), vertices (0, 3/2) and (0, -5/2), endpoints (1, -1/2) and (-1, -1/2), foci (0, -1/2 ±√3), eccentricity √3/2.

Step 1: Rewrite the Equation in Standard Form

The given equation is \(4x^2 + (y + \frac{1}{2})^2 = 4\). We need to rewrite this in the standard form of an ellipse. Divide each term by 4 to get \(\frac{x^2}{1} + \frac{(y + \frac{1}{2})^2}{4} = 1\). This corresponds to the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \(a^2 = 1\) and \(b^2 = 4\).

Step 2: Identify the Center

The equation \(\frac{(x-0)^2}{1} + \frac{(y + \frac{1}{2})^2}{4} = 1\) indicates the center is at \((h, k) = (0, -\frac{1}{2})\).

Step 3: Determine the Vertices and Orientation

Since \(b > a\) (\(b^2 = 4\) and \(a^2 = 1\)), the major axis is vertical. The vertices are at \(\left(0, -\frac{1}{2} \pm 2\right)\), which are \(\left(0, \frac{3}{2}\right)\) and \(\left(0, -\frac{5}{2}\right)\).

Step 4: Find the Endpoints of the Minor Axis

The endpoints of the minor axis are located horizontally from the center. They are at \((0 \pm 1, -\frac{1}{2})\), which gives the points \((1, -\frac{1}{2})\) and \((-1, -\frac{1}{2})\).

Step 5: Calculate the Foci

To find the foci, calculate \(c\) using the formula \(c = \sqrt{b^2 - a^2}\). Here, \(c = \sqrt{4 - 1} = \sqrt{3}\). The foci are along the major axis, located at \((0, -\frac{1}{2} \pm \sqrt{3})\).

Step 6: Determine Eccentricity

The eccentricity \(e\) is given by \(e = \frac{c}{b}\). Therefore, \(e = \frac{\sqrt{3}}{2}\) because \(b = 2\).

Step 7: Graph the Ellipse

To graph the ellipse, plot the center at \((0, -\frac{1}{2})\), the vertices at \((0, \frac{3}{2})\) and \((0, -\frac{5}{2})\), the minor axis endpoints at \((1, -\frac{1}{2})\) and \((-1, -\frac{1}{2})\), and the foci at \((0, -\frac{1}{2} \pm \sqrt{3})\). Connect these points with an ellipse-shaped curve.

Key Concepts

Center of an Ellipse

The center of an ellipse plays a crucial role as it acts as the midpoint around which the ellipse is symmetrically shaped. To determine the center from the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), you simply look for the values of \(h\) and \(k\), which are the coordinates of the center.
In our standard form equation, \(\frac{(x-0)^2}{1} + \frac{(y + \frac{1}{2})^2}{4} = 1\), the center is located at \((h, k) = (0, -\frac{1}{2})\).
This is because the terms \(h\) and \(k\) are the constants accompanied by \(x\) and \(y\) in the transformed parts of their respective equations.
The center tells us where to start when sketching or analyzing an ellipse on a graph.

Foci of an Ellipse

The foci are two specific points inside the ellipse that have an important geometric property. Their locations define the shape of the ellipse. In mathematical terms, any point on the ellipse is such that the sum of the distances to the two foci is constant.
To find the foci for an ellipse, you need to calculate \(c\), which is given by the formula \[c = \sqrt{b^2 - a^2}\]. From this, the foci are positioned along the major axis relative to the center of the ellipse.
In our example, since the major axis is vertical, the foci numbers are \((0, -\frac{1}{2} \pm \sqrt{3})\).
Remember, finding foci helps understand the ellipse's stretch and orientation, showing how the ellipse deviates from being a circle.

Vertices of an Ellipse

The vertices are the extremes, or farthest points, on an ellipse along the major axis. There are always two vertices in any ellipse, and they help highlight the true extent of the ellipse's size and orientation on its major axis.
For our ellipse, identified by \(\frac{(x-0)^2}{1} + \frac{(y+\frac{1}{2})^2}{4} = 1\), since \(b^2 > a^2\), it tells us the major axis is vertical and therefore the vertices lie in the vertical direction from the center.
The vertices' formula are located at \((h, k \pm b)\). Substituting in our ellipse's center, we find the vertices at \((0, \frac{3}{2})\) and \((0, -\frac{5}{2})\).
Knowing the vertices not only helps in sketching the ellipse but also provides full insight into its vertical stretch.

Eccentricity of an Ellipse

Eccentricity allows us to quantify the deviation of an ellipse from a perfect circle. It is a positive number that is less than 1, uniquely defining how "stretched" an ellipse is.
Given by the formula \(e = \frac{c}{b}\), where \(c\) is the distance calculated from the center to each focus and \(b\) is the semi-major axis length.
For our ellipse, \(c = \sqrt{3}\) and \(b = 2\), thus \(e = \frac{\sqrt{3}}{2}\).
A smaller eccentricity means the ellipse is more circular, while a larger value implies a more elongated shape. Understanding eccentricity helps in comparing different ellipses and understanding the scale of stretch in relation to a circle.