Problem 2 - Algebra and Trigonometry

Question

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: (6.87, -3) and (-0.87, -3). Major axis: 8, Minor axis: 2.

Step 1: Identify the Standard Form

The equation of the ellipse is given as \( \frac{(x-3)^2}{16} + (y+3)^2 = 1 \). This is in the standard form of \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \( (h, k) \) is the center of the ellipse, \( a \) is the semi-major axis, and \( b \) is the semi-minor axis. Since there is no denominator under the \( (y+3)^2 \) term, it can be rewritten as \( \frac{(y+3)^2}{1} \), identifying \( b^2 = 1 \).

Step 2: Determine the Center

From the standard form, \( (h, k) = (3, -3) \). Thus, the center of the ellipse is \((3, -3)\).

Step 3: Identify the Major and Minor Axes Lengths

Since \( a^2 = 16 \), we calculate \( a = 4 \) and \( b^2 = 1 \) gives \( b = 1 \). Therefore, the lengths of the major and minor axes are \( 2a = 8 \) and \( 2b = 2 \), respectively. The larger value corresponds to \( a \, (\frac{(x-3)^2}{16})\), indicating the ellipse is horizontally oriented.

Step 4: Calculate the Vertices

The vertices along the major axis are \((h \pm a, k)\), so we find \((3 \pm 4, -3)\), which yields the vertices at \((7, -3)\) and \((-1, -3)\).

Step 5: Calculate the Foci

For an ellipse, the foci are found using \( c = \sqrt{a^2 - b^2} \). Calculate \( c = \sqrt{16 - 1} = \sqrt{15} \). Thus, the foci are at \((h \pm c, k) = (3 \pm \sqrt{15}, -3)\), approximately \((3 + 3.87, -3)\) and \((3 - 3.87, -3)\) or \((6.87, -3)\) and \((-0.87, -3)\).

Step 6: Sketch the Ellipse

To sketch the ellipse, draw a horizontal oval centered at \((3, -3)\) with the major axis extending 4 units in each direction horizontally and the minor axis extending 1 unit in each direction vertically. Mark the vertices and foci on the graph as calculated.

Key Concepts

Center of an Ellipse

The center of an ellipse is the very center point, or origin, around which the ellipse is symmetrical. In the standard form of \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \] this center is represented by the ordered pair \((h, k)\). In our given equation, \[ \frac{(x-3)^2}{16} + (y+3)^2 = 1, \] the center is identified by closely following the numbers accompanying the \(x\) and \(y\) coordinates.

The formula reveals that the center of this ellipse is located at \((3, -3)\).

• The number 3 comes from the \((x-3)^2\) part, showing a horizontal shift to the right.
• Similarly, the number -3 comes from the \((y+3)^2\) part, indicating a downward vertical shift.

Always remember, the center of an ellipse helps to position the ellipse correctly on the coordinate plane.

Foci of an Ellipse

The foci of an ellipse are two key points located along the major axis. These points help in defining the overall shape and "flatness" of the ellipse. For any ellipse, the sum of the distances from both foci to any point on the ellipse is constant.

To find the distance from the center to each focus, use the equation \( c = \sqrt{a^2 - b^2} \). Based on our specific equation, we have \( a^2 = 16 \) and \( b^2 = 1 \).

• Calculate \( c = \sqrt{16 - 1} = \sqrt{15} \).
• This gives us an approximate value of \( c \approx 3.87 \).
• The foci are then situated at \((h \pm c, k)\), which becomes \((3 \pm 3.87, -3)\). This delivers the points \((6.87, -3)\) and \((-0.87, -3)\).

The foci are essential in understanding the elliptical shape and are always found on the major axis, equidistant from the center.

Vertices of an Ellipse

The vertices of an ellipse are pivotal points located at the maximum distance from the center on the major axis. These define the longest dimension of the ellipse. To find the vertices in a horizontally-oriented ellipse, use the formula \((h \pm a, k)\).

In our equation, \( a = 4 \), so the vertices are found by adding and subtracting this value from the \(h\)-coordinate.

• The vertices become \((3 + 4, -3)\) and \((3 - 4, -3)\).
• The specific coordinates are then \((7, -3)\) and \((-1, -3)\).

These vertices mark the "extremes" of the ellipse along the major axis and are crucial in aiding the visual representation of the ellipse.

Major and Minor Axes

The major and minor axes tell us about the orientation and proportions of an ellipse. The major axis is the longest diameter of the ellipse, and the minor axis is the shortest.

• The length of the major axis is determined by \(2a\).
• The length of the minor axis is \(2b\).

In our equation, the major axis is horizontal since \( a > b \) that is, \( a = 4 \) and with \( a^2 = 16 \).

• Thus, the length of the major axis is \(2 \times 4 = 8\).
• The minor axis has \( b = 1 \) and the length is \(2 \times 1 = 2\).

Understanding these dimensions helps to graph the ellipse correctly, giving insight into its orientation and elongation.

Ellipse Sketching

Sketching an ellipse involves translating the algebraic information into a visual form. This process is aided immensely by understanding the elliptical elements like the center, vertices, foci, and axes you've discovered.

Start with the center at \((3, -3)\). From this point, you can sketch your ellipse.

• First, draw the major axis, extending 4 units horizontally to each side from the center.
• Next, sketch the minor axis, reaching 1 unit vertically in both directions.
• Plot the vertices at \((7,-3)\) and \((-1, -3)\).
• Also, ensure to mark the foci at \((6.87, -3)\) and \((-0.87, -3)\).

By joining the end points of these axes smoothly, the shape of the ellipse emerges.This detailed representation provides clarity on how different essentially calculated elements of the ellipse interact.