Problem 55 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

Center: (-1, -1). Vertices: (-2, -1), (0, -1), (-1, -3), (-1, 1).

Step 1: Rearrange and Group Terms

Rearrange the terms in the equation by grouping the x-terms together and the y-terms together. Start with the equation: \[ 4x^2 + 8x + y^2 + 2y + 1 = 0 \] Move constant to the other side: \[ 4x^2 + 8x + y^2 + 2y = -1 \]

Step 2: Complete the Square for x

First, factor out the 4 from the x-terms:\[ 4(x^2 + 2x) + y^2 + 2y = -1 \]To complete the square inside the bracket: Take half of the coefficient of x (which is 2), square it (1), and add and subtract inside the brackets:\[ 4((x^2 + 2x + 1) - 1) + y^2 + 2y = -1 \] Simplify to:\[ 4((x+1)^2 - 1) + y^2 + 2y = -1 \] Distribute the 4:\[ 4(x+1)^2 - 4 + y^2 + 2y = -1 \]

Step 3: Complete the Square for y

Complete the square for the y-terms in a similar way. Take half of 2, which is 1, square it (1), and add and subtract:\[ 4(x+1)^2 - 4 + (y^2 + 2y + 1 - 1) = -1 \] This gives:\[ 4(x+1)^2 - 4 + (y+1)^2 - 1 = -1 \]

Step 4: Simplify the Equation

Combine and move constants to the other side of the equation:\[ 4(x+1)^2 + (y+1)^2 = 4 \]Divide every term by 4 to normalize the equation:\[ \frac{4(x+1)^2}{4} + \frac{(y+1)^2}{4} = \frac{4}{4} \]Simplifying gives the standard form:\[ (x+1)^2 + \frac{(y+1)^2}{4} = 1 \]

Step 5: Identify the Center and Vertices

The equation is now in the standard form for an ellipse: \[ \left( \frac{x-h}{a} \right)^2 + \left( \frac{y-k}{b} \right)^2 = 1 \] where (h, k) is the center, a² is the denominator of the x-term, and b² is the denominator of the y-term. Here, a² = 1 and b² = 4. Thus, the center is (-1, -1), a = 1, so vertex (along x-axis) at (-2, -1) and (0, -1), b = 2, so vertex (along y-axis) at (-1, -3) and (-1, 1).

Key Concepts

Standard Form of Ellipse

An ellipse is a special kind of curve on a plane that forms a closed loop, characterized by its two axes: the major axis and the minor axis. The standard form of an ellipse's equation is \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where:

• \((h, k)\) is the center of the ellipse.
• \(a\) is the distance from the center to the vertices along the x-axis.
• \(b\) is the distance from the center to the vertices along the y-axis.

If \(a > b\), the major axis is horizontal; if \(b > a\), it's vertical.
To identify the standard form, it's crucial to manipulate the given equation through techniques such as completing the square.
Doing so helps to express the quadratic terms in a way that clearly shows the center and lengths of the axes.

Completing the Square

Completing the square is a technique used in algebra to make expressions into perfect square trinomials. This technique is especially useful in rewriting the equation of an ellipse into its standard form.
The main idea is to take a quadratic expression, like \(ax^2 + bx\), and transform it into \((x-h)^2\), where \(h\) is part of the perfect square. To complete the square:

• Start with the quadratic: \(x^2 + bx\).
• Find half of the coefficient of \(x\), which is \(\frac{b}{2}\).
• Square it: \(\left( \frac{b}{2} \right)^2\).
• Add and subtract this square inside the expression.

This process rewrites the expression into a perfect square trinomial, making it easier to see the translation involved (the \(h\) and \(k\) shift) as well as the overall shape of the ellipse.

Center of Ellipse

The center of an ellipse is a crucial point, as it serves as a reference for the location and orientation of the entire curve. In the standard form of an ellipse, the center is represented by the coordinates \((h, k)\).
These coordinates are simply the values that slide the ellipse from the origin to its correct placement in the coordinate plane. To find the center:

• When the equation is in its standard form \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\), \(h\) is the horizontal shift, and \(k\) is the vertical shift.
• The expression \((x-h)\) moves the ellipse right or left.
• The expression \((y-k)\) moves it up or down.

Understanding the center's coordinates \((-1, -1)\) from our exercise tells us about the directional displacement from the origin for the ellipse. Knowing the center is essential for locating other features, such as vertices and foci.