Problem 19 - College Algebra

Question

For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. $$ 4 x^{2}-8 x+9 y^{2}-72 y+112=0 $$

Step-by-Step Solution

Answer

The ellipse is \(\frac{(x-1)^2}{9} + \frac{(y-4)^2}{4} = 1\). Endpoints are \((4,4), (-2,4)\), \((1,6), (1,2)\). Foci are \((1\pm\sqrt{5},4)\).

Step 1: Write the Given Equation

Begin with the given equation of the ellipse:\[4x^2 - 8x + 9y^2 - 72y + 112 = 0\]

Step 2: Group and Complete the Square for x-terms

Group the x-terms together and complete the square:\[4(x^2 - 2x)\]Complete the square for \(x^2 - 2x\):\(x^2 - 2x = (x-1)^2 - 1\)So,\[4[(x-1)^2 - 1] = 4(x-1)^2 - 4\]

Step 3: Group and Complete the Square for y-terms

Group the y-terms together and complete the square:\[9(y^2 - 8y)\]Complete the square for \(y^2 - 8y\):\(y^2 - 8y = (y-4)^2 - 16\)So,\[9[(y-4)^2 - 16] = 9(y-4)^2 - 144\]

Step 4: Rewrite the Equation with Completed Squares

Substitute the completed squares back into the equation and rearrange:\[4(x-1)^2 - 4 + 9(y-4)^2 - 144 + 112 = 0\]Combine the constants:\[4(x-1)^2 + 9(y-4)^2 - 36 = 0\]

Step 5: Convert into Standard Form

Add 36 to both sides to get the equation in standard form for ellipses:\[4(x-1)^2 + 9(y-4)^2 = 36\]Divide all terms by 36:\[\frac{(x-1)^2}{9} + \frac{(y-4)^2}{4} = 1\]

Step 6: Identify Key Features

The center of the ellipse is at \((1, 4)\).The semi-major axis is along the x-axis since 9 (under the x-term) is greater than 4.**Major Axis:** 2a, where \(a = 3\) (since \(a^2 = 9\)).Endpoints are \((1+3, 4)\) and \((1-3, 4)\) which are \((4, 4)\) and \((-2, 4)\).**Minor Axis:** 2b, where \(b = 2\) (since \(b^2 = 4\)).Endpoints are \((1, 4+2)\) and \((1, 4-2)\) which are \((1, 6)\) and \((1, 2)\).**Foci:** Calculate c using \(c^2 = a^2 - b^2\), so \(c = \sqrt{5}\).Foci are \((1+\sqrt{5}, 4)\) and \((1-\sqrt{5}, 4)\).

Key Concepts

Standard Form of an Ellipse

Ellipses are a type of conic section that can be represented by a specific algebraic equation known as the standard form. In general, the standard form of an ellipse equation is:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here,

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

In more practical terms, the standard form provides us an organized way to understand the orientation and scaling of the ellipse. If \(a^2\) is greater than \(b^2\), the major axis is parallel to the x-axis. Conversely, if \(b^2\) is greater than \(a^2\), the major axis is parallel to the y-axis. In the solved problem, the given equation was rearranged into this form:\[ \frac{(x-1)^2}{9} + \frac{ (y-4)^2}{4} = 1 \]This indicates the ellipse is centered at \((1, 4)\). The standard form gives clear visibility into its attributes, simplifying calculations about the ellipse's dimensions and orientation.

Understanding the Major Axis

The major axis of an ellipse is the longest diameter cutting through its center. This axis provides essential insights into the ellipse's overall size. In the standard form equation, the term with the larger denominator indicates the direction of the major axis. For the example provided:

• The term with the larger denominator is under \((x-1)^2\), which is 9.
• This tells us that the major axis is along the x-axis.

The length of the major axis is calculated as \(2a\), where \(a=\sqrt{9}=3\). This means the endpoints of the major axis occur at 3 units from the center along the x-direction. Thus, the endpoints are at - \((1+3, 4)\) or \((4, 4)\)- \((1-3, 4)\) or \((-2, 4)\). Recognizing the major axis helps visualize and sketch the ellipse correctly, providing a baseline for further effective analysis.

Understanding the Minor Axis

While the major axis extends through the longest part of the ellipse, the minor axis serves as the shorter diameter, perpendicular to the major axis. Observing the standard form of the ellipse allows us to find this minor dimension easily. In the scenario examined,

• The term with the smaller denominator is under \((y-4)^2\), which is 4.
• Hence, the minor axis runs parallel to the y-axis.

The length of the minor axis is given by \(2b\), where \(b=\sqrt{4}=2\). Therefore, the endpoints are 2 units from the center along the y-direction. The minor axis endpoints are located at:- \((1, 4+2)\) or \((1, 6)\)- \((1, 4-2)\) or \((1, 2)\). Understanding where the minor axis lies provides further insight into the dimensions of the ellipse and confirms its proportions visually.

Locating the Foci

In the fascinating geometry of ellipses, the foci are two important points situated along the major axis. The total distance from any point on the ellipse to both foci is constant, giving ellipses their unique shape. To determine the foci, we use the relationship \[ c^2 = a^2 - b^2 \].In this exercise:

• \(a^2 = 9\) and \(b^2 = 4\), so \(c = \sqrt{5}\).

Since the ellipse is oriented with the major axis along the x-axis, the foci are aligned horizontally as relative to the center.Thus, the foci are at:- \((1 + \sqrt{5}, 4)\)- \((1 - \sqrt{5}, 4)\). These calculated positions help describe the stretch and the "ovalness" of the ellipse, underpinning its geometric peculiarity, and showcasing the role of foci in the definition of ellipses as a curve.