Problem 39 - A Graphical Approach to College Algebra

Question

Write the equation in standard form for an ellipse centered at ( \(h, k\) ). Identify the center and vertices. $$9 x^{2}+18 x+4 y^{2}-8 y-23=0$$

Step-by-Step Solution

Answer

Center: (-1, 1), Vertices: (-1, 4) and (-1, -2).

Step 1: Rearrange the Equation

Begin by rearranging the equation, grouping the x terms and y terms together:\[ 9x^2 + 18x + 4y^2 - 8y = 23 \] This moves the constant term to the right side of the equation.

Step 2: Complete the Square for x Terms

Complete the square for the x terms. Factor out 9 from the x terms:\[ 9(x^2 + 2x) \]Take half of the coefficient of x (2), square it (1), and add it inside the parenthesis:\[ 9(x^2 + 2x + 1) - 9 \times 1 \]Add and subtract \(9 \times 1 \) outside the parentheses to keep the equation balanced, modifying the rearranged equation to:\[ 9(x + 1)^2 + 4y^2 - 8y = 23 + 9 \] resulting in:\[ 9(x + 1)^2 \] which modifies the constant on the right to 32.

Step 3: Complete the Square for y Terms

Similarly, complete the square for the y terms. Factor out 4 from the y terms:\[ 4(y^2 - 2y) \]Take half of the coefficient of y (-2), square it (1), and add it inside the parenthesis:\[ 4(y^2 - 2y + 1) - 4 \times 1 \]Add and subtract \(4 \times 1 \) outside the parentheses, modifying term on the right:\[ 4(y - 1)^2 \] which further alters the constant on the right to 36.

Step 4: Rewrite the Equation in Standard Form

Balance the modified constant terms and express the ellipse equation:\[ 9(x + 1)^2 + 4(y - 1)^2 = 36 \]Divide the entire equation by 36 to set the right-hand side to 1:\[ \frac{(x + 1)^2}{4} + \frac{(y - 1)^2}{9} = 1 \]This is the equation of the ellipse in standard form.

Step 5: Identify the Center and Vertices

Identify the center and vertices from the standard form equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \):- Center is \((-1, 1)\).- For the vertices, since \(a^2=4\) and \(b^2=9\), vertices are vertically aligned with the center at \((-1, 1\pm3)\), thus vertices are \((-1, 4)\) and \((-1, -2)\).

Key Concepts

Standard Form

The standard form for the equation of an ellipse is essential when understanding its geometry. An ellipse can be described by the following equation:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here,

• \((h, k)\) is the center of the ellipse.
• \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

In this form, one can easily identify the ellipse's position and orientation. The right side of the equation is always equal to 1. This equation showcases the distance of all points on the ellipse from the center and is vital for finding the center and lengths of semi-axes.

When the ellipse is oriented horizontally, \(a > b\). If oriented vertically, \(b > a\). Transforming an equation into this form allows you to quickly gather vital information about the ellipse.

Completing the Square

Completing the square is a method used to transform quadratic equations into a more manageable form. This is crucial when converting a general conic section equation into the standard form of an ellipse. By grouping and manipulating the terms, you can rewrite parts of the equation to reveal a perfect square trinomial.

For the exercise, we applied this technique as follows:

• Group the \(x\) terms and \(y\) terms separately, maintaining balance by keeping constants on the other side.
• Factor out the leading coefficient of the quadratic terms to simplify completing the square.
• Add and subtract the same value inside the parentheses to maintain the identity of the equation. This involves taking half of the linear coefficient and squaring it.

The result is a quadratic expression in the form \((x+d)^2\), where \(d\) is calculated to complete the square.

This process helps reveal the structure necessary to put the equation in standard form.

Equation of Ellipse

An ellipse's equation is a compact way to represent all points that satisfy the distance criteria from a fixed center. After completing the square and arranging terms, the standard form is achieved. Let's focus on an example: \[\frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1\]This equation is divided by 36 to get 1 on the right side, ensuring the standard form is fulfilled.

Important aspects to remember are:

• The terms in the equation represent squared binomials normalized by the square of the lengths of the axes.
• This illustrates how the ellipse changes by adjusting \(a\) and \(b\); larger denominators indicate longer axes.

Thus, understanding this equation helps in visualizing the ellipse's geometry, including centricity and axes proportions. The transformation into this form also highlights alterations necessary when dealing with real-world applications.

Center and Vertices

Identifying the center and vertices of an ellipse is straightforward once in standard form. From the example equation, \(\frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1\), we derive important characteristics:

• The center \((h, k)\) is found at \((-1, 1)\).
• The distances from the center to the vertices give insight into its shape and size: \(a = 2\) and \(b = 3\).

Since \(b > a\), the ellipse is vertically oriented.

The vertices follow vertically from the center:

• By adding and subtracting \(b\) to/from the \(y\) component: \((-1, 1\pm3)\).
• This yields the vertices \((-1, 4)\) and \((-1, -2)\).

This understanding not only confirms proper form conversion but also aids in drawing and interpreting ellipses visually and mathematically.