Problem 6 - Algebra and Trigonometry

Question

\(5-8\) . Find the center, foci, and vertices of the ellipse, and determine the lengths of the major and minor axes. Then sketch the graph. $$ \frac{(x-3)^{2}}{16}+(y+3)^{2}=1 $$

Step-by-Step Solution

Answer

Center: (3, -3), Vertices: (7, -3) and (-1, -3), Foci: (3±√15, -3). Major axis: 8, minor axis: 2.

Step 1: Identify the Standard Form

The equation is given as \(\frac{(x-3)^{2}}{16}+(y+3)^{2}=1\). This resembles the standard form of an ellipse \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\), where \((h,k)\) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

Step 2: Identify the Center

From the equation \(\frac{(x-3)^{2}}{16} + (y+3)^{2} = 1\), we identify \(h = 3\) and \(k = -3\). Thus, the center of the ellipse is \((3, -3)\).

Step 3: Determine Semi-Axes and Orientation

Here, \(a^2 = 16\) and \(b^2 = 1\). Hence, \(a = 4\) and \(b = 1\). The ellipse is oriented horizontally because \(a > b\). The major axis is along the x-axis.

Step 4: Calculate Lengths of the Axes

The major axis length is \(2a = 8\), and the minor axis length is \(2b = 2\).

Step 5: Find Vertices

The vertices are located along the major axis at \((h \pm a, k)\). So, the vertices are at \((3 \pm 4, -3)\) or \((7, -3)\) and \((-1, -3)\).

Step 6: Find Foci

The focal distance \(c\) is found using the formula \(c^2 = a^2 - b^2\). Substituting the values, we get \(c^2 = 16 - 1 = 15\), thus \(c = \sqrt{15}\). The foci are located at \((h \pm c, k)\), or \((3 \pm \sqrt{15}, -3)\).

Step 7: Sketch the Graph

Plot the center \((3, -3)\). Draw the major axis horizontally from \(-1\) to \(7\) and the minor axis vertically from \((3, -4)\) to \((3, -2)\). Plot the vertices and foci to sketch the ellipse.

Key Concepts

Ellipse Equation

An ellipse is a geometric shape that can be described by an algebraic equation. The given ellipse equation is \(\frac{(x-3)^{2}}{16}+(y+3)^{2}=1\). This equation represents an ellipse with a horizontal orientation.
The equation of an ellipse can be compared to the standard form \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\). Here, \((h, k)\) denotes the center of the ellipse, and \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively.
In the provided equation, you can observe that \(a^2 = 16\) and \(b^2 = 1\). The larger denominator under the \((x-h)^2\) term suggests the ellipse is wider than it is tall, indicating a horizontal major axis. Understanding how to identify these elements within the equation is key to solving and graphing ellipses.

Standard Form of Ellipse

The standard form of the ellipse equation is vital for understanding how the ellipse looks and where it's located. Recalling the equation, \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\), allows us to easily pinpoint attributes such as the center and the directions of the axes.
For the exercise, the given equation, \(\frac{(x-3)^{2}}{16}+(y+3)^{2}=1\), directly falls into the standard form. Here:

• \(h = 3\) and \(k = -3\) – indicating that the center of the ellipse is at \((3, -3)\).
• \(a^2 = 16\) implying \(a = 4\) – meaning the horizontal semi-major axis is 4 units long.
• \(b^2 = 1\) meaning \(b = 1\) – indicating that the vertical semi-minor axis is 1 unit long.

The standard form provides a simple yet powerful method to visualize and solve problems involving ellipses, especially for finding various points like the center or the foci.

Ellipse Center

The center of an ellipse is an important feature often depicted with the coordinates \((h, k)\) in the standard form. From our example equation, \(\frac{(x-3)^{2}}{16}+(y+3)^{2}=1\), identifying \(h = 3\) and \(k = -3\) was straightforward by matching the equation's components to its standard form structure.
This means the center of the ellipse is located at \((3, -3)\). This point acts as a pivot for the ellipse, allowing us to determine the positions of other crucial parts like the vertices and foci. Knowing the center's position helps in sketching and understanding the geometric properties of the ellipse.

Foci and Vertices of Ellipse

The foci and vertices are key features of an ellipse that help describe its geometry. They can be quickly found once the ellipse is broken down into its standard form.
For our ellipse equation, the vertices are derived from the centers and semi-major axis length. With \(a = 4\), the vertices are positioned along the major axis at \((3 \pm 4, -3)\), which calculates to \((7, -3)\) and \((-1, -3)\).
The foci require the calculation of the value \(c\) using \(c^2 = a^2 - b^2\), which results in \(c = \sqrt{15}\). The foci are then placed at \((3 \pm \sqrt{15}, -3)\).
Vertices indicate the farthest points on the ellipse along the major axis, while the foci are the points from which the sum of the distances to any point on the ellipse remains constant. Recognizing the positions of these can greatly aid in graphing and interpreting ellipses.