Problem 58 - 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. $$ 2 x^{2}+4 x+3 y^{2}-18 y+23=0 $$

Step-by-Step Solution

Answer

The ellipse is centered at (-1, 3) with vertices extending \(a=\sqrt{3}\) horizontally and \(b=\sqrt{2}\) vertically.

Step 1: Group and Prepare Terms

The given equation is \(2x^2 + 4x + 3y^2 - 18y + 23 = 0\). First, group the \(x\) terms and \(y\) terms together for completion of the square. This gives us: \(2(x^2 + 2x) + 3(y^2 - 6y) = -23\).

Step 2: Complete the Square for x-terms

To complete the square for \(x\), take half of the coefficient of \(x\), square it and add inside the parenthesis. Coefficient of \(x\) is 2, so \((2/2)^2 = 1\). Add and subtract 1 inside the \(x\) terms: \(2((x^2 + 2x + 1) - 1)\). This simplifies to \(2((x+1)^2 - 1)\).

Step 3: Complete the Square for y-terms

Now, complete the square for \(y\) terms. The coefficient of \(y\) is -6, so \((-6/2)^2 = 9\). Add and subtract 9 inside the \(y\) terms: \(3((y^2 - 6y + 9) - 9)\). Simplify to \(3((y-3)^2 - 9)\).

Step 4: Substitute the Completion back and Simplify

Substitute the completed squares back into the equation: \(2[(x+1)^2 - 1] + 3[(y-3)^2 - 9] = -23\). Distribute and simplify: \(2(x+1)^2 - 2 + 3(y-3)^2 - 27 = -23\), leading to \(2(x+1)^2 + 3(y-3)^2 = 6\).

Step 5: Divide by the Simplified Constant

Divide the entire equation by 6 to express it in standard form: \(\frac{2(x+1)^2}{6} + \frac{3(y-3)^2}{6} = 1\). Simplify: \(\frac{(x+1)^2}{3} + \frac{(y-3)^2}{2} = 1\).

Step 6: Identify the Center and Vertices

The ellipse is in the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) with \(h = -1\), \(k = 3\), \(a^2 = 3\), and \(b^2 = 2\). The center is at \((-1, 3)\). The vertices are determined by the distances \(a = \sqrt{3}\) along the x-axis and \(b = \sqrt{2}\) along the y-axis.

Key Concepts

Standard Form of an Ellipse Equation

The standard form of an ellipse's equation provides a neat and organized way to capture all essential properties of an ellipse. It's generally presented as \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
This equation tells us:

• \( (h, k) \) is the center of the ellipse.
• \( a^2 \) and \( b^2 \) are the denominators that define the width and height of the ellipse, or in other words, the distances from the center to the vertices along the x and y axes.

The "1" on the right side of the equation helps us identify it as an ellipse in its standard form. Understanding this form is crucial because it allows us to instantly see where the center of the ellipse is and how far it stretches in both directions.

Completing the Square

Completing the square is a technique used to reshape parts of a quadratic equation into something more useful and simpler to analyze, particularly in ellipse equations. Here is how it works:

• For any quadratic expression like \( x^2 + bx \), you find half of \( b \), square it, and add it back.
• This creates a perfect square trinomial, \( (x + \frac{b}{2})^2 \).

Applying this to both \( x \) and \( y \) terms in an equation helps transform it into the standard ellipse form. The process involves:

• Grouping like terms.
• Performing completion of the square on each group separately.
• Simplifying the equation to fit the standard ellipse form.

This method turns complex expressions into a simple and insightful shape.

Ellipse Center

The center of an ellipse is a fundamental part of its equation. It is denoted by \((h, k)\) in the standard form equation. It acts as the midpoint and the balance point of the ellipse, helping to distribute its shape evenly around that central point. In the conversion example we worked with, by identifying terms, the completed equation was \( \frac{(x+1)^2}{3} + \frac{(y-3)^2}{2} = 1 \).
Thus, the center of the ellipse is at \((-1, 3)\). Knowing the center is essential for other calculations like distances and relationships within the ellipse.

Ellipse Vertices

Vertices of an ellipse provide us with critical information about its size and orientation. They lie directly on the principal axes, extending outwards from the center. To find the vertices:

• Calculate the distances \(a = \sqrt{a^2}\) and \(b = \sqrt{b^2}\).
• For our example, \(a = \sqrt{3}\), and \(b = \sqrt{2}\).

These distances are then used to calculate the points of the vertices along the x and y axes starting from the center.
For the ellipse centered at \((-1, 3)\):

• The vertices along the x-axis are at \((-1 \pm \sqrt{3}, 3)\).
• The vertices along the y-axis are at \((-1, 3 \pm \sqrt{2})\).

This gives a clear image of the ellipse's extent in each direction.