Problem 53 - 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. $$ 9 x^{2}+18 x+4 y^{2}-8 y-23=0 $$

Step-by-Step Solution

Answer

Center: (-1, 1); Vertices: (-3, 1), (1, 1), (-1, -2), (-1, 4).

Step 1: Rearrange the Equation

Start by grouping the terms involving each variable together, and move the constant to the right side of the equation:\[ 9x^2 + 18x + 4y^2 - 8y = 23 \]

Step 2: Complete the Square for x-terms

For the \(x\) terms, factor out 9 and complete the square:\[ 9(x^2 + 2x) \]To complete the square, add and subtract \(1\) inside the parenthesis:\[ 9((x + 1)^2 - 1) = 9(x + 1)^2 - 9 \]

Step 3: Complete the Square for y-terms

For the \(y\) terms, factor out 4 and complete the square:\[ 4(y^2 - 2y) \]To complete the square, add and subtract \(1\) inside the parenthesis:\[ 4((y - 1)^2 - 1) = 4(y - 1)^2 - 4 \]

Step 4: Arrange the Completed Squares

Substitute the completed squares back into the equation:\[ 9(x+1)^2 - 9 + 4(y-1)^2 - 4 = 23 \]

Step 5: Simplify the Equation

Combine all constants on the right side:\[ 9(x+1)^2 + 4(y-1)^2 = 36 \]

Step 6: Divide by 36 and Write in Standard Form

Divide the entire equation by 36 to convert it to the standard form for an ellipse:\[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1 \]

Step 7: Identify the Center and Vertices

The center of the ellipse is at \((h, k) = (-1, 1)\).The vertices along the \(x\)-axis are at \((-1 \pm 2, 1)\), which are \((-3, 1)\) and \((1, 1)\).The vertices along the \(y\)-axis are at \((-1, 1 \pm 3)\), which are \((-1, -2)\) and \((-1, 4)\).

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This method is crucial for converting the general form of a conic section equation into its standard form. Let's break down this process, specifically applied to finding the standard form of an ellipse.

To complete the square for the expression involving a variable, follow these steps:

• Identify the terms to focus on—for instance, consider the equation in terms of \(x\).
• Factor out the coefficient of the quadratic term. In the exercise, for \(9x^2 + 18x\), factor out 9, leading to \(9(x^2 + 2x)\).
• Take half of the linear term's coefficient (i.e., the coefficient of \(x\), which is 2 in this case), square it, and add & subtract it inside the parenthesis. So, \((2/2)^2 = 1\) is added and subtracted: \((x + 1)^2 - 1\).
• Adjust the constant outside by distributing, resulting in \(9(x + 1)^2 - 9\).

The same technique is applied to the \(y\) terms, gathering \(4(y^2 - 2y)\) and completing the square similarly. Completing the square helps structurally reorganize the equation to easily identify the components that form the ellipse's geometric representation.

Center of an Ellipse

Determining the center of an ellipse is essential in understanding its position on the coordinate plane. The standard form of an ellipse is \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where \((h, k)\) represents the center of the ellipse.

In the given exercise, after rearranging and completing the square, the equation transforms to\[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1 \]Looking closely, the expression \((x+1)^2\) indicates a horizontal shift to \(x = -1\). Similarly, \((y-1)^2\) signifies a vertical shift to \(y = 1\). Therefore, the center of the ellipse becomes \((-1, 1)\).

Identifying the center by completing squares is straightforward because it directly follows from the transformation of the variables \(x\) and \(y\). This step assures that all parts of the ellipse equation are correctly aligned geometrically, serving as a foundation to measure distances to the vertices and foci.

Vertices of an Ellipse

Finding the vertices of an ellipse helps in understanding its spread on the coordinate system. An ellipse's vertices lie along the direction of its major and minor axes. Let's deduce them with simple reasoning.

For the standard form of the ellipse equation:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]\(a\) and \(b\) denote the distances from the center to the vertices along the principal axes. In the equation from the exercise \[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1 \]the terms \(4 = a^2\) and \(9 = b^2\) tell us about these critical distances.

From this, we see:

• The horizontal distance \(a = 2\) units from the center \((-1, 1)\), hence the vertices along the x-axis are \((-1 - 2, 1)\) and \((-1 + 2, 1)\), simplified to \((-3, 1)\) and \((1, 1)\).
• The vertical distance \(b = 3\) units, makes the vertices along the y-axis \((-1, 1 - 3)\) and \((-1, 1 + 3)\), simplified to \((-1, -2)\) and \((-1, 4)\).

Knowing the location of vertices adds to the geometric understanding, completing the visualization of how data points spread symmetrically around the center.