Problem 2 - College Algebra

Question

Graph each ellipse and locate the foci. $$ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 $$

Step-by-Step Solution

Answer

The center of the ellipse is at (0,0). The vertices are at (±5,0) and co-vertices are at (0, ±4). The foci are at (±3, 0).

Step 1: Identify the center

The center of the ellipse is at the origin (0,0) because the equation is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Step 2: Locate the vertices and co-vertices

The vertices of the ellipse are at (±5,0) because \(a^2 = 25\). Hence, a = 5. Similarly, the co-vertices are at (0, ±4) because \(b^2 = 16\). Therefore, b = 4.

Step 3: Compute the foci

The foci of the ellipse are at (±c, 0) where \(c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = ±3\). So the foci are at (±3, 0).

Step 4: Draw the ellipse

Plot the center, vertices, co-vertices, and two foci on a graph. Draw a smooth curve connecting the points to form the ellipse.

Key Concepts

Ellipse Equation

Understanding the ellipse equation is fundamental to graphing an ellipse. An ellipse is a set of points where the sum of the distances from two fixed points (the foci) to any point on the ellipse is constant. The standard form of the equation for an ellipse centered at the origin is \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \(a\) is the distance from the center to the vertex on the major axis and \(b\) is the distance to the co-vertex on the minor axis. When \(a > b\), the ellipse is stretched along the x-axis, and when \(b > a\), it's stretched along the y-axis.

Let's break it down further; the ellipse equation captures the proportional relationship between any point \((x,y)\) on the ellipse and the distances \(a\) and \(b\):

• If \(x=0\), then \(y\) must be either \(b\) or \(-b\) which are the endpoints on the minor axis.
• If \(y=0\), then \(x\) must be either \(a\) or \(-a\) which are the endpoints on the major axis.

These points mark the vertices and co-vertices, pivotal in sketching the basic shape before fine-tuning with other details like the foci.

Vertices of an Ellipse

The vertices of an ellipse are perhaps the most prominent features. They are the points farthest from the center along the major axis. For our example, the equation \[\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\] indicates the vertices are at \((±5,0)\).

Vertices give the ellipse its longest diameter, the major axis, which helps determine its shape and orientation. They are essential for drawing the ellipse as they set the boundary in the direction of the major axis. In practical graphing steps, identifying the vertices first involves:

• Finding the value of \(a\) by taking the square root of the larger denominator in the ellipse equation.
• Plotting the points \((±a,0)\) relative to the center when the major axis is horizontal.

A proper understanding of the vertices also aids in concepts like eccentricity, where the closeness of the vertices to the foci defines the 'flatness' of the ellipse.

Foci of an Ellipse

The foci (singular: focus) are two fixed points inside the ellipse equidistant from the center along the major axis. They are central to the ellipse's definition and have a unique property; the sum of the distances from the foci to any point on the ellipse's edge is constant. For our example, we calculate the foci as \((±3,0)\).

We find the foci by using the formula \(c = \sqrt{a^2 - b^2}\), where \(c\) is the distance from the center to each focus. In a graph, plotting the foci involves:

• Calculating the value of \(c\) as described.
• Marking the points \((±c,0)\), if the major axis is horizontal.

The foci play a vital role when dealing with the reflective properties of ellipses and their application in areas like optics and acoustics. It's fascinating how the design of whispering galleries and satellite dishes use the principle where sound or signals are reflected off the surface to focus at one of the foci.