Problem 38 - Precalculus

Question

Find the center, vertices, foci, and eccentricity of the ellipse. Then sketch the ellipse. $$\frac{(x+3)^{2}}{12}+\frac{(y-2)^{2}}{16}=1$$

Step-by-Step Solution

Answer

Center: (-3, 2), Vertices: (-3±2√3, 2), Foci: (-5, 2) and (-1, 2), Eccentricity: √3/3

Step 1: Identify the Center

Center (h,k) of an ellipse is given by the values in the equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\). In the given equation \(\frac{(x+3)^2}{12}+\frac{(y-2)^2}{16}=1\), the center is at (-3, 2).

Step 2: Find the Vertices

The vertices are located at a units to the left and right, and b units above and below the center. From the given ellipse equation, we can identify that \(a^2=12\) so \(a=\sqrt{12}=2\sqrt{3}\), and \(b^2=16\) so \(b=4\). Hair, the vertices are at (-3±a, 2), so (-3±2\sqrt{3}, 2).

Step 3: Calculate the Foci

The foci of an ellipse are located on the major axis, c units from the center, where \(c=\sqrt{a^2-b^2}\). From our values, \(c=\sqrt{12-16} = \sqrt{4} = 2\). Therefore, the foci are at (-3±c, 2), so (-5, 2) and (-1, 2).

Step 4: Determine the Eccentricity

Eccentricity is the ratio of the distance between the foci and the length of the major axis, given by \(e=\frac{c}{a}\). Substituting our values gives \(e=\frac{2}{2\sqrt{3}}=\frac{\sqrt{3}}{3}\).

Step 5: Sketch the Ellipse

By plotting the center, vertices and foci on a coordinate plane, the sketch of the ellipse can be created. The center is at (-3, 2), the vertices are at (-3±2\sqrt{3}, 2), and the foci are at (-5, 2) and (-1, 2). Lines are drawn to connect these points in an elliptical shape.

Key Concepts

Center of the Ellipse

The center of an ellipse is a significant point of reference for defining its overall shape and position. It provides the middle point around which the ellipse is perfectly symmetric. Understanding how to find the center is crucial.

• An ellipse's equation in standard form is given by \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
• Here, \( (h,k) \) represents the center of the ellipse.
• In our particular equation, \[ \frac{(x+3)^2}{12} + \frac{(y-2)^2}{16} = 1 \]we replace \( h \) with -3 and \( k \) with 2.

Thus, the center of this ellipse is located at the point (-3, 2). The center helps us establish the location of other components like vertices and foci.

Vertices of the Ellipse

Vertices are the farthest points on an ellipse along its major axis. These points are essential for gauging the width of an ellipse and understanding its orientation.

• For an ellipse, the vertices can be found using \( a \) and \( b \), which are the semi-major and semi-minor axes.
• Specifically for our problem, \[ a^2 = 12 \Rightarrow a = \sqrt{12} = 2\sqrt{3} \] \[ b^2 = 16 \Rightarrow b = 4 \]

The vertices will be horizontally displaced from the center \( (h,k) \).

• They are located at points \((h \pm a, k)\)for this horizontal ellipse.
• Thus, the vertices are: \[ (-3 \pm 2\sqrt{3}, 2) \]

So for our particular ellipse, the vertices lie at positions approximately at \( (-3 + 2\sqrt{3}, 2) \) and \( (-3 - 2\sqrt{3}, 2) \). These points dictate the ellipse’s maximal spread horizontally.

Foci of the Ellipse

The foci, or singular focus, are key points on the major axis of an ellipse. They determine the shape's eccentricity and render the ellipse more defined.

• In the context of an ellipse, the foci always lie along the major axis, and their location can be identified by computing \( c \).
• The formula for determining \( c \) is \[ c^2 = a^2 - b^2 \]
• Applying this, we have:\[ c^2 = 12 - 16 = 4 \Rightarrow c = \sqrt{4} = 2 \]

Thus, the foci are located at positions \((h \pm c, k)\) for a horizontal ellipse, which here gives \[ (-3 \pm 2, 2) \]

Specifically, these translate to the coordinates (-5, 2) and (-1, 2). The foci not only inform the directional extent of the ellipse but also connect to the concept of eccentricity.

Eccentricity of the Ellipse

Eccentricity is a measure of how much an ellipse deviate from a perfect circle. It ranges from 0 to 1, where circles have an eccentricity of 0. The closer the value is to 1, the more elongated the ellipse looks.

• For any ellipse, eccentricity \( e \) can be calculated by the formula\[ e = \frac{c}{a} \]
• Inserting the values determined earlier for our ellipse gives:\[ e = \frac{2}{2\sqrt{3}} = \frac{\sqrt{3}}{3} \]

This fractional value emphasizes that our ellipse is elongated but not excessively so.

Understanding eccentricity provides insight into the shape's quality and helps visualize how narrow or broad an ellipse is. Recognizing the eccentricity in relation to the other properties of the ellipse aids in sketching and contextualizing its properties in real-world applications.