Problem 55 - Algebra 2

Question

Find the vertices and co-vertices of each ellipse. $$ \frac{x^{2}}{34}+\frac{y^{2}}{25}=1 $$

Step-by-Step Solution

Answer

The vertices of the ellipse are at (-5.83, 0) and (5.83, 0) while the co-vertices are at (0, -5) and (0, 5).

Step 1: Identify the semi-major and semi-minor axes

The given equation is \(\frac{x^{2}}{34}+\frac{y^{2}}{25}=1\). Here, 34 under \(x^{2}\) is greater than 25 under \(y^2\), which means a^2=34 and b^2=25. We find the square root to obtain the lengths: \(a=\sqrt{34}\) and \(b=\sqrt{25}\). Then \(a\approx5.83\) and \(b=5\). So the longer axis of the ellipse is along the x-direction (horizontal), and this is the major axis (length of 2a). The shorter axis is along the y-direction (vertical) and this is the minor axis (length of 2b).

Step 2: Locate the vertices

The vertices of an ellipse are at \(\pm a\) along the major axis. Since the major axis is along the x-direction, the vertices are at \((\pm a, 0) = (\pm 5.83, 0) = (-5.83, 0) , (5.83, 0)\).

Step 3: Identify the co-vertices

The co-vertices of an ellipse are at \(\pm b\) along the minor axis. Since the minor axis is along the y-direction, the co-vertices are at \((0, \pm b) = (0, \pm 5) = (0, -5), (0, 5)\).

Key Concepts

Vertices of an Ellipse

In the study of ellipses, the vertices are a crucial component to understand. Vertices are the points on an ellipse that are closest and farthest from the center along its major axis. They determine the overall 'length' of the ellipse and are used to define the stretch in one of the principal directions. The major axis is the longest diameter that runs through the center and touches the ellipse at both vertices.

The given ellipse equation is \(\frac{x^{2}}{34}+\frac{y^{2}}{25}=1\). From this equation, the number under \(x^2\) is larger, or 34, which means it defines the semi-major axis, so \(a^2=34\). When you take the square root, you get \(a=\sqrt{34}\approx 5.83\). This indicates the major axis is horizontal.

Therefore, the vertices of this ellipse are located at \((\pm 5.83, 0)\), that is

• \((-5.83, 0)\)
• \((5.83, 0)\)

These are the points where the ellipse is the widest along the x-axis.

Co-vertices of an Ellipse

The concept of co-vertices complements the vertices of an ellipse. Co-vertices are found along the minor axis, which is the shortest diameter through the ellipse. The minor axis runs perpendicular to the major axis and provides a measurement of how "tall" or "narrow" the ellipse is.

In our equation \(\frac{x^{2}}{34}+\frac{y^{2}}{25}=1\), the value 25 under the \(y^2\) term represents \(b^2\), where \(b\) is the semi-minor axis. Taking the square root provides \(b=\sqrt{25}=5\). This indicates a vertical minor axis, meaning the ellipses' narrowness is top to bottom.

The co-vertices are then positioned at \((0, \pm 5)\), specifically as:

• \((0, -5)\)
• \((0, 5)\)

These points are vertically above and below the center of the ellipse.

Semi-Major Axis

Understanding the semi-major axis is fundamental in grasping how ellipses are structured. This axis is half the length of the longest diameter passing through the center of the ellipse, known as the major axis. The semi-major axis extends from the center to one vertex. If you imagine the ellipse as a stretched circle, the semi-major axis defines this horizontal stretching.

In the equation \(\frac{x^{2}}{34}+\frac{y^{2}}{25}=1\), the larger number under the \(x^2\) term is 34, indicating that \(a^2\) is equal to 34 and \(a = \sqrt{34} \approx 5.83\). This means the semi-major axis is approximately 5.83 units long.
The major axis of the ellipse, therefore, spans horizontally from the far-left vertex \((-5.83, 0)\) to the far-right vertex \((5.83, 0)\).

When examining ellipses, it's important to note the relationship this has with the semi-minor axis, as both define the full scale and dimension of the ellipse.