Problem 58 - Algebra and Trigonometry

Question

In Exercises \(51-60,\) convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$ 49 x^{2}+16 y^{2}+98 x-64 y-671=0 $$

Step-by-Step Solution

Answer

The equation in standard form becomes \[\frac{(x+1)^{2}}{14}+\frac{(y-1/2)^{2}}{42.5}=1\]. The center of the ellipse is \((-1, 1/2)\), lengths of the axes are \(2\sqrt{14}\) and \(2\sqrt{42.5}\) respectively. The foci are located at \((-1, 1/2 \pm \sqrt{14})\).

Step 1: Convert to Standard Form

First, rearrange the equation by grouping the \(x\) terms together and \(y\) terms together and move the constant to the right side of the equation \[ 49(x^{2}+2x)+16(y^{2}-4y) = 671 \]Now, complete the square for the \(x\) and \(y\) terms to achieve the standard form of an ellipse equation. \[ 49[(x+1)^{2}-1]+16[(y-\frac{1}{2})^{2}-(\frac{1}{2})^{2}] = 671 \]Simplify the equation to obtain the standard form, which should resemble the format \((x-h)^{2}/a^{2}+(y-k)^{2}/b^{2}=1\), where \((h,k)\) are the center coordinates of the ellipse.\[ 49(x+1)^{2}+16(y-\frac{1}{2})^{2} = 680 \]Divide the whole equation by 680 to get a 1 on the right side\[ \frac{(x+1)^{2}}{14}+\frac{(y-1/2)^{2}}{42.5}=1\]

Step 2: Determine Ellipse Characteristics

After finding standard form, use it to find the center, lengths of the major \(2a\) and minor \(2b\) axes, and foci of the ellipse. The center is formed by the values within the parentheses multiplied by negative. In this case, the center is at \((-1,1/2)\). The lengths of the minor and major axes are \(2\sqrt{14}\) and \(2\sqrt{42.5}\). The foci of the ellipse can be found by using the formula \(c=\sqrt{|a^2-b^2|}\), where \(c\) is the distance from the center to each focus, \(a\) is the length from the center of the ellipse to the end of the major axis, and \(b\) is the length from the center to the end of the minor axis, in this case, \(c = \sqrt{14}\). The location of the foci will be \((-1, 1/2 \pm \sqrt{14})\)

Step 3: Graph the Ellipse

To graph the ellipse, start by plotting the center of the ellipse. Next, plot the ends of the major axis and minor axis based on the values found. Draw a smooth curve to connect these points to form the ellipse. Then, place the foci at their calculated locations

Key Concepts

Completing the Square

Completing the square is a method used to rewrite a quadratic expression in a form that reveals important geometric information, like the center of an ellipse. To complete the square for a quadratic polynomial, you rearrange terms and add and subtract the same value internally to create a perfect square trinomial.

For example, take the expression from the equation:

• First, group the x terms together and the y terms separately, like this: \[49(x^2+2x) + 16(y^2-4y) = 671\]
• Then, complete the square on each group: \[49((x+1)^2 - 1) + 16((y - \frac{1}{2})^2 - (\frac{1}{2})^2) = 671\]

This rearrangement helps to transform the equation into the standard form of an ellipse, which is crucial for identifying its properties.

Standard Form of an Ellipse

The standard form of an ellipse provides a structured framework to easily identify its geometric properties. Once you've completed the square, this form looks like:

• \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

Here,

• \(h\) and \(k\) represent the coordinates of the center of the ellipse.
• \(a\) and \(b\) are the lengths from the center to the vertices of the ellipse along the major and minor axes, respectively.

To achieve this form, divide through by the constant to have a "1" on the right side:

• \[\frac{(x+1)^2}{14} + \frac{(y-\frac{1}{2})^2}{42.5} = 1\]

This standard form directly reveals the ellipse's center at \((-1, \frac{1}{2})\) and helps in understanding the orientation and dimensions of the ellipse.

Ellipse Foci

The foci (plural of focus) of an ellipse are two special points located along its major axis. Understanding the foci is essential since they represent the focal characteristic of ellipses, where the sum of the distances from each point on the ellipse to both foci remains constant.

To find the foci, use the formula for the distance from the center to each focus, \(c\):

• \[c = \sqrt{|a^2 - b^2|}\]

In our exercise, \(a\) and \(b\) are the respective half-lengths of the major and minor axes: \(a^2 = 42.5\) and \(b^2 = 14\). Thus,

• \[c = \sqrt{42.5 - 14} = \sqrt{28.5}\]

So, the foci are positioned at \((-1, \frac{1}{2} \pm \sqrt{28.5})\), residing along the major axis's vertical alignment in our case study.

Graphing Ellipses

Graphing ellipses involves plotting key points and then sketching the oval shape. Begin with:

• Locate the center of the ellipse using \((h, k)\) found during standard form transformation, which is \((-1, \frac{1}{2})\) for this example.
• Plot the vertices along the major and minor axes. Calculate these using \(a\) and \(b\) to know their distance from the center:

For our ellipse with \(a = \sqrt{42.5}\) and \(b = \sqrt{14}\):

• Plot points at \((-1, \frac{1}{2} \pm \sqrt{42.5})\) for the vertical major axis.
• Place at \((-1 \pm \sqrt{14}, \frac{1}{2})\) for the horizontal minor axis.

Draw a smooth curve through these points to create the ellipse. Finally, plot the foci found previously to aid in understanding the ellipse's geometric structure.