Problem 20 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ x^{2}+3 y^{2}+18 y+18=0 $$

Step-by-Step Solution

Answer

Center: (0, -3); Foci: (±√6, -3); Vertices: (±3, -3); Minor Axis Endpoints: (0, -3±√3); Eccentricity: √2/3. Graph with horizontal major axis.

Step 1: Rewrite the Equation

We begin by rewriting the given equation for clarity:\[ x^{2}+3y^{2}+18y+18=0 \]This equation implies we need to complete the square for the \( y \) terms.

Step 2: Complete the Square for the y-Terms

First, isolate the \( y \) terms:\[ 3(y^2 + 6y) = -x^2 - 18 \]Now, complete the square for \( y^2 + 6y \):1. Take half of 6 (the coefficient of \( y \)), square it to get 9.2. Add and subtract 9 inside the bracket to complete the square:\[ 3((y+3)^2 - 9) = -x^2 - 18 \]Simplify to:\[ 3(y+3)^2 - 27 = -x^2 - 18 \]

Step 3: Rearrange into Ellipse Form

Rewrite the equation with completed squares:\[ 3(y+3)^2 = -x^2 - 18 + 27 \]\[ 3(y+3)^2 = -x^2 + 9 \]Divide the entire equation by \(-9\) to standardize:\[ \frac{x^2}{9} + \frac{(y+3)^2}{3} = 1 \]

Step 4: Identify the Ellipse Elements

From the standardized form, \( \frac{x^2}{9} + \frac{(y+3)^2}{3} = 1 \), identify the elements:- Center: (0, -3) (from \( x^2 \) and \( (y+3)^2 \))- Major axis (horizontal) with \( a^2 = 9 \) so \( a = 3 \)- Minor axis has \( b^2 = 3 \) so \( b = \sqrt{3} \)- Eccentricity \( e = \frac{c}{a} \) where \( c^2 = a^2 - b^2 = 9 - 3 = 6 \), so \( c = \sqrt{6} \).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a form that is easier to work with. This technique is particularly useful when dealing with conic sections like ellipses, hyperbolas, and parabolas. In our original equation, \( x^2 + 3y^2 + 18y + 18 = 0 \), it is necessary to complete the square for the \( y \) terms to simplify the equation.

Here's how you do it:

• Isolate the \( y \) terms, which gives: \( 3(y^2 + 6y) = -x^2 - 18 \)
• Take half of the coefficient of \( y \) (which is 6), square it to get 9.
• Add and subtract 9 within the bracket: \( 3((y+3)^2 - 9) = -x^2 - 18 \)

This transforms the quadratic \( y \) terms into a perfect square, making the equation more manageable for further operations.

Standard Form of Ellipse

The standard form of an ellipse helps identify its key properties more easily. After completing the square and rearranging the terms, the original equation \( x^2 + 3y^2 + 18y + 18 = 0 \) was rewritten into its standardized form as \( \frac{x^2}{9} + \frac{(y+3)^2}{3} = 1 \).

This setup reveals the following:

• The center of the ellipse is \((0, -3)\), indicating a shift from the origin.
• The denominators, 9 and 3, represent \( a^2 \) and \( b^2 \) respectively, demonstrating the lengths of the semi-major and semi-minor axes. Here, \( a = 3 \) and \( b = \sqrt{3} \).

The standardized form simplifies graphing and allows easy identification of the ellipse's elements.

Eccentricity

Eccentricity is a key characteristic of conic sections, representing how much an ellipse deviates from being circular. For ellipses, eccentricity \( e \) ranges between 0 and 1. A value closer to 0 signifies a more circular shape.

In our case, to find the eccentricity of the derived ellipse equation \( \frac{x^2}{9} + \frac{(y+3)^2}{3} = 1 \), follow these steps:

• Use the formula \( c^2 = a^2 - b^2 \) to find \( c \). Plug in the values to get \( c^2 = 9 - 3 = 6 \), so \( c = \sqrt{6} \).
• Calculate the eccentricity using \( e = \frac{c}{a} = \frac{\sqrt{6}}{3} \).

This measurement helps describe the shape and spread of the ellipse's foci relative to its center.

Ellipse Elements

Identifying the elements of an ellipse involves understanding its key characteristics such as center, foci, vertices, and axes.

From the equation \( \frac{x^2}{9} + \frac{(y+3)^2}{3} = 1 \), we gather:

• The center is at \((0, -3)\).
• The major axis is horizontal with a length of \( 2a = 6 \), with vertices at \((3, -3)\) and \((-3, -3)\).
• The minor axis is vertical with a length of \( 2b = 2\sqrt{3} \), with endpoints at \((0, -3+\sqrt{3})\) and \((0, -3-\sqrt{3})\).
• The foci are positioned along the major axis, calculated using \( c = \sqrt{6} \), located at \((\sqrt{6}, -3)\) and \((-\sqrt{6}, -3)\).

Recognizing these elements eases the process of plotting the ellipse and aids visual comprehension of its structure.