Problem 39 - 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. $$9 x^{2}+18 x+4 y^{2}-8 y-23=0$$

Step-by-Step Solution

Answer

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

Step 1: Rearrange the Equation

To begin, rearrange the terms of the given equation to group the terms with the same variables. Rearrange it as follows:\[ 9x^2 + 18x + 4y^2 - 8y = 23. \]

Step 2: Complete the Square for x-terms

Take the x-terms, \(9x^2 + 18x\), and factor out the 9:\[ 9(x^2 + 2x). \]To complete the square, take half of the coefficient of \(x\), which is 2, divide by 2 to get 1, and square it to get 1. Add and subtract 1 inside the parenthesis:\[ 9(x^2 + 2x + 1 - 1) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9. \]

Step 3: Complete the Square for y-terms

Now, consider the y-terms \(4y^2 - 8y\), and factor out the 4:\[ 4(y^2 - 2y). \]For completing the square, take half of -2, which is -1, square it to get 1. Add and subtract 1 inside the parenthesis:\[ 4(y^2 - 2y + 1 - 1) = 4((y-1)^2 - 1) = 4(y-1)^2 - 4. \]

Step 4: Combine and Simplify

Combine the completed square terms with the rearranged equation:\[ 9(x+1)^2 - 9 + 4(y-1)^2 - 4 = 23. \]Simplify by combining constant terms:\[ 9(x+1)^2 + 4(y-1)^2 = 36. \]

Step 5: Divide by 36 to Get Standard Form

To convert the equation into standard form, divide every term by 36:\[ \frac{9(x+1)^2}{36} + \frac{4(y-1)^2}{36} = \frac{36}{36}. \]Simplify each fraction:\[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1. \]Now the equation is in the standard form of an ellipse.

Step 6: Identify Center and Vertices

The standard form of the equation is \(\frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1\), which is an ellipse centered at \((-1,1)\).For the ellipse, \(a^2 = 9\) and \(b^2 = 4\), hence \(a = 3\) and \(b = 2\).The vertices are at \((-1, 1+3) = (-1, 4)\) and \((-1, 1-3) = (-1, -2)\).

Key Concepts

Completing the Square

To understand the process of completing the square, let's start with the general idea. Completing the square is a method used to transform a quadratic expression, making it easier to solve or simplify.
In the context of ellipses, it helps to reshape the equation into a standard form.

Here's how to complete the square step-by-step:

• Begin with the quadratic expression, for example: \( ax^2 + bx + c \).
• Factor out the coefficient of \( x^2 \) if it is not 1 or divide the entire equation by that coefficient to simplify.
• Take half of the coefficient of \( x \) (after factoring if necessary), square it, then add and subtract this square within the equation group.

In our case, for the \( x \)-terms \( 9x^2 + 18x \), we first factor out the 9, leading to an expression of \( 9(x^2 + 2x) \).
Taking half of 2 gives 1, squaring it results in 1, so we add and subtract 1 inside the parentheses:

• \( 9(x^2 + 2x + 1 - 1) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9 \).

Similarly, follow the same approach for the \( y \)-terms. After factoring out the 4 from \( 4y^2 - 8y \), complete the square:

• \( 4(y^2 - 2y + 1 - 1) = 4((y-1)^2 - 1) = 4(y-1)^2 - 4 \).

This technique simplifies the quadratic into a perfect square trinomial, which is vital for writing the ellipse equation in its standard form.

Standard Form of a Conic Section

The standard form of an ellipse equation provides a simple way to identify its key features, such as its center, vertices, and axes lengths. This form is derived after completing the square and offers clearer insights into the ellipse's structure.

The standard equation of an ellipse centered at \((h, k)\) looks like this:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \].
Here:

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

For a horizontal ellipse, \(a > b\), and for a vertical ellipse, \(b > a\).

In our example, after completing the square, the equation becomes:
\[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1 \].
The center of the ellipse is \((-1, 1)\), and the lengths of the semi-major and semi-minor axes are derived from \(a^2 = 9\) and \(b^2 = 4\).
This provides essential information to visualize the ellipse and assess its dimensions.

Center and Vertices of an Ellipse

The center and vertices are essential components in graphing and understanding an ellipse's geometry. Knowing how to determine these points allows for a complete analysis of the ellipse's shape and position.

The center of an ellipse is given in its standard form equation as \((h, k)\). This point marks the midpoint along which the ellipse is symmetric. In our exercise, the center is located at \((-1, 1)\).

Next, the vertices of the ellipse indicate the points where the ellipse reaches its maximum extent along each axis. The vertices are positioned \(a\) units away from the center along the major axis.
Given the equation:
\[ \frac{(x+1)^2}{4} + \frac{(y-1)^2}{9} = 1 \],
we identify \(a^2 = 9\) and \(b^2 = 4\), so \(a = 3\) and \(b = 2\).
Since the larger value \(a\) corresponds with the vertical terms, the major axis is vertical.

The vertices are then calculated as:

• \((-1, 1 + 3) = (-1, 4)\)
• \((-1, 1 - 3) = (-1, -2)\)

By pinpointing these features, one can estimate the overall size and shape of the ellipse, enhancing the ability to graph it accurately.