Problem 41 - A Graphical Approach to Precalculus with Limits

Question

Write the equation in standard form for an ellipse centered at ( \(h, k\) ). Identify the center and vertices. $$4 x^{2}+8 x+y^{2}+2 y+1=0$$

Step-by-Step Solution

Answer

The ellipse is centered at (-1, -1) with vertices at (-1, 1) and (-1, -3).

Step 1: Group and Reorder Terms

Start by rearranging the equation to group the x-terms and y-terms separately:\[4x^2 + 8x + y^2 + 2y + 1 = 0\] becomes \[(4x^2 + 8x) + (y^2 + 2y) = -1\].

Step 2: Complete the Square for x-Terms

For the x-terms, \(4x^2 + 8x\), factor out the 4: \[4(x^2 + 2x)\].To complete the square inside the parentheses, take half of the linear coefficient (2), square it, and add and subtract this square inside the parenthesis: \[x^2 + 2x + 1 - 1\].Thus the equation becomes:\[4((x + 1)^2 - 1)\].

Step 3: Complete the Square for y-Terms

Next, complete the square for \(y^2 + 2y\).Take half of 2 (1), square it (1), and add and subtract it: \[y^2 + 2y + 1 - 1 = (y+1)^2 - 1\].

Step 4: Simplify Equation

Plug the completed squares back into the equation:\[4((x + 1)^2 - 1) + (y + 1)^2 - 1 = -1\].Simplify:\[4(x + 1)^2 - 4 + (y + 1)^2 - 1 = -1\].\[4(x + 1)^2 + (y + 1)^2 = 4\].

Step 5: Divide to Achieve Standard Form

Divide the whole equation by 4 to match the standard form of an ellipse:\[\frac{4(x+1)^2}{4} + \frac{(y+1)^2}{4} = \frac{4}{4}\],simplifying to:\[(x+1)^2 + \frac{(y+1)^2}{4} = 1\].This is the equation in standard form for an ellipse.

Step 6: Identify the Center and Vertices

The center of the ellipse is at \((-1, -1)\). The semi-major axis is vertical (because 4 is larger than 1), and \(b = \sqrt{4} = 2\). The semi-minor axis is horizontal with \(a = 1\). The vertices are at \((-1, -1+2)\), \((-1, -1-2)\). So, vertices are \((-1, 1)\) and \((-1, -3)\).

Key Concepts

Completing the Square

The technique of "completing the square" is crucial when working with conic sections like ellipses. It transforms quadratic expressions into a form that is easier to interpret and manipulate. This method involves making a perfect square trinomial from a quadratic expression.

• Start by grouping terms with the same variable.
• Factor out any coefficients in front of the squared term. For example, in the expression \(4x^2 + 8x\), factor out the 4: \(4(x^2 + 2x)\).
• Next, take half of the linear coefficient (in this case, the coefficient of \(x\)), square it, and both add and subtract this term inside the parenthesis: \(x^2 + 2x + 1 - 1\).
• This transforms the expression into a squared binomial: \( (x + 1)^2 - 1 \).

Repeat these steps for each variable group to complete the square for both \(x\) and \(y\) terms. Once completed, you can simplify and rearrange the equation to set the stage for identifying ellipse components.

Standard Form of Ellipse

The standard form of the ellipse's equation reveals much about its geometric properties. This form is pivotal for graphing and understanding the ellipse's layout.

• For an ellipse, the standard form is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) are the coordinates of the center.
• In the exercise, after completing the square, the equation became \((x+1)^2 + \frac{(y+1)^2}{4} = 1\).
• Notice how each term is divided by a number, ensuring the equation equals 1. This form aids in identifying the lengths of the ellipse’s axes.

The transformation highlights the ellipse's orientation and dimensions. Always ensure the equation equals 1; this is imperative for identifying other components.

Identifying Ellipse Components

Understanding the completed and simplified equation helps in identifying key components of the ellipse.
For a ellipse, these components include:

• The equation \((x+1)^2 + \frac{(y+1)^2}{4} = 1\) tells us that the squared terms’ denominators are crucial.
• The values beneath the terms \((x+1)^2\) and \((y+1)^2\) are \(a^2\) and \(b^2\), respectively.
• In this example: \(a^2 = 1\) and \(b^2 = 4\). Hence, \(a = 1\) which gives us the length of the semi-minor axis and \(b = 2\) which gives us the length of the semi-major axis.

This differentiation between \(a\) and \(b\) signifies the direction and length of the ellipse's axes, crucial for accurately graphing the ellipse.

Ellipse Center and Vertices

The center and the vertices of an ellipse are the vital points needed for its graphing.

• The center of an ellipse, found easily when in standard form, is \((h, k)\).
• For our ellipse, the center is located at -1, -1, derived directly from the terms \((x+1)\) and \((y+1)\). Always change the signs of these values.
• Vertices are found along the axes determined by the lengths \(a\) and \(b\).
• Given \(a = 1\) and \(b = 2\), and knowing that \(b > a\), the major axis is vertical.
• Add and subtract \(b\) from the y-coordinate of the center to find the vertices: -1, -1 + 2) and -1, -1 - 2).
• Resulting in vertices at -1, 1) and -3).

Understanding the placement of the center and vertices provides clarity in plotting the ellipse correctly on a graph.