Problem 9 - Precalculus: Functions and Graphs

Question

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$\frac{(x-3)^{2}}{16}+\frac{(y+4)^{2}}{9}=1$$

Step-by-Step Solution

Answer

Vertices: (7, -4), (-1, -4); Foci: (3+√7, -4), (3-√7, -4).

Step 1: Identify Equation Form

The given equation is \(\frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^2}{b^2} = 1\). Since \(16 > 9\), \(a^2 = 16\) and \(b^2 = 9\). The center of the ellipse is \((h, k) = (3, -4)\).

Step 2: Calculate Lengths of Semi-Major and Semi-Minor Axes

Since \(a^2 = 16\), \(a = \sqrt{16} = 4\), and \(b^2 = 9\), \(b = \sqrt{9} = 3\). This means that the semi-major axis is 4 and the semi-minor axis is 3. The major axis is horizontal because \(a > b\).

Step 3: Determine Vertices

Vertices along the major axis are horizontally aligned. Thus, they are located \(a\) units to the left and right of the center. Therefore, the vertices are \((3 + 4, -4) = (7, -4)\) and \((3 - 4, -4) = (-1, -4)\).

Step 4: Calculate Foci

The distance from the center to each focus \(c\) is given by \(c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}\). The foci are located \(c\) units to the left and right of the center along the major axis. Thus, the foci are \((3 + \sqrt{7}, -4)\) and \((3 - \sqrt{7}, -4)\).

Step 5: Sketch the Ellipse

Start by plotting the center at \((3, -4)\). Then, plot the vertices \((7, -4)\) and \((-1, -4)\), and co-vertices \((3, -7)\) and \((3, -1)\). Plot the foci \((3 + \sqrt{7}, -4)\) and \((3 - \sqrt{7}, -4)\). Draw the ellipse through these points, ensuring it is elongated horizontally.

Key Concepts

Vertices of an ellipse

The vertices of an ellipse are crucial points that help define its shape and position. By definition, the vertices of an ellipse are located along its major axis. In the context of the given equation \(\frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1\), the major axis is horizontal. To find the vertices, you identify the distance from the center to the vertices using the semi-major axis length, which is denoted by \(a\).
For this ellipse, the center is at \((3, -4)\) and the length \(a\) is 4. This means the vertices will be 4 units away from the center along the x-axis.

• The right vertex is \((3 + 4, -4) = (7, -4)\).
• The left vertex is \((3 - 4, -4) = (-1, -4)\).

These vertices are key points to sketch an accurate graph of the ellipse and indicate its significant width.

Foci of an ellipse

The foci (plural of "focus") are two points located along the major axis and play a critical role in the geometric properties of an ellipse. The distance from any point on the ellipse to one focus plus its distance to the other focus remains constant.
To find the foci for the ellipse \(\frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1\), you can use the formula for the distance from the center to each focus \(c\): \(c = \sqrt{a^2 - b^2}\).
Here, \(a^2=16\) and \(b^2=9\). Calculate \(c\) as follows:
\[c = \sqrt{16 - 9} = \sqrt{7}\]
The foci are found by moving \(\sqrt{7}\) units from the center along the major axis.

• The right focus is \((3 + \sqrt{7}, -4)\).
• The left focus is \((3 - \sqrt{7}, -4)\).

Understanding these points is vital for both accurately graphing the ellipse and for deeper exploration of its properties in geometry.

Semi-major and semi-minor axes

An ellipse has two critical axes which determine its overall dimensions: the semi-major and semi-minor axes. These axes are perpendicular to each other and intersect at the ellipse's center.
The semi-major axis is the longer one, while the semi-minor is the shorter. Their lengths help define the shape's elongation and orientation.
In the given ellipse equation \(\frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1\), the term under the larger denominator (16) corresponds to the semi-major axis. Thus, \(a^2 = 16\), leading to \(a = \sqrt{16} = 4\). For the semi-minor axis, \(b^2 = 9\), so \(b = \sqrt{9} = 3\).
The lengths are:

• Semi-major axis: 4 units
• Semi-minor axis: 3 units

Because the semi-major axis length is greater, the ellipse stretches horizontally. These axes' lengths will aid you in accurately scaling and plotting the ellipse's graph.