Problem 44 - College Algebra with Corequisite Support

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(100 x^{2}+1000 x+y^{2}-10 y+2425=0\)

Step-by-Step Solution

Answer

The ellipse has center (-5, 5), vertices at (-7.5, 5) and (-2.5, 5), and foci at (-5, 29.845) and (-5, -19.845).

Step 1: Rearrange the equation

Let's start by rearranging the terms of the given equation by grouping the terms involving \(x\) and \(y\) together: \[100x^2 + 1000x + y^2 - 10y = -2425\]

Step 2: Complete the square for x and y terms

Now, we'll complete the square for both the \(x\) and \(y\) terms. Starting with the \(x\) terms:- Extract the common factor of 100 from \(x\): \[100(x^2 + 10x)\]- Use \(10/2 = 5\), then square to get 25. Add and subtract 25 inside the bracket:\[100((x^2 + 10x + 25) - 25) = 100((x+5)^2 - 25)\]- Now for the \(y\) terms:- Use \(5/2 = 2.5\) then square to get 6.25. Add and subtract it:\[(y^2 - 10y + 25) - 6.25 = (y-5)^2 - 25\]Now, substitute back into the equation:\[100((x+5)^2 - 25) + (y-5)^2 - 25 = -2425\]

Step 3: Simplify and transform into standard form

Distribute and combine constants:- For \(x\): \[100(x+5)^2 - 2500\]- The combined form: \[100(x+5)^2 + (y-5)^2 - 2500 - 25 = -2425\]- Combine numerical terms: \[100(x+5)^2 + (y-5)^2 = 0\]Divide the entire equation by 625 (since both terms need to be simplified by assuming a circle and fixing for ellipses):\[\frac{100(x+5)^2}{625} + \frac{(y-5)^2}{625} = 1\]Simplify again to fully see the ellipse's standard form:\[\frac{(x+5)^2}{6.25} + \frac{(y-5)^2}{625} = 1\]This is essentially a horizontal ellipse.

Step 4: Identify elements of the ellipse

From the simplified standard form:\[\frac{(x+5)^2}{6.25} + \frac{(y-5)^2}{625} = 1\]- Center: \((-5, 5)\)- Vertices along the x-axis: From the term \(\frac{(x+5)^2}{6.25}\), the distance is \(\sqrt{6.25} = 2.5\), thus vertices at \((-7.5, 5)\) and \((-2.5, 5)\)- Major axis length is related to \(\sqrt{625} = 25\) along y-axis, hence foci are calculated using \(c^2 = b^2 - a^2\) where \(a = 25\) and \(b = 2.5\), placing them at same \(x=-5\) in intervals of \(\sqrt{625-6.25}=24.845\) in the y-axis up and down from center.

Key Concepts

Completing the Square for Ellipses

Completing the square is a fundamental method used in converting quadratic equations into a more recognizable and analyzable form. This technique is crucial when dealing with ellipse equations, as it helps rewrite the general form equation into the standard form. Here, we break down its application step-by-step. To "complete the square," start by rearranging the initial equation so that terms involving the same variable are grouped together. For example, let's look at the equation that involves both \(x\) and \(y\) terms: \[100x^2 + 1000x + y^2 - 10y = -2425\]The idea is to make each grouped term into a perfect square trinomial. This involves:

• Extracting common factors to simplify the expression.
• Adding and subtracting inside the brackets to form binomials squared. For \(x\), we take \(x^2 + 10x\) and find the number which completes the square.
• Repeat a similar process for any terms involving \(y\).

This process forms a set of square binomials, transforming the quadratic equation into the standard ellipse form.

Graphing Ellipses

Once the equation of the ellipse is in its standard form, graphing becomes more intuitive and straightforward. The standard form is typically expressed as: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here, \((h, k)\) denotes the center of the ellipse, and \(a\) and \(b\) represent the distances from the center to the vertex along the x-axis and y-axis, respectively. The general steps to graph this shape include:

• Identifying the center \((h, k)\) of the ellipse from the standard form.
• Plotting the center on a graph is your starting point.
• Using values of \(a\) and \(b\), determine the vertices by moving \(a\) units along the x-axis and \(b\) units along the y-axis from the center, noting these new points.
• Sketch the major and minor axes, which pass through the center and vertices.
• Draw the elliptical curve, maintaining smoothness and symmetry around these axes.

By following these steps, you create an accurate visual representation of the ellipse, enhancing understanding of its spatial properties.

Ellipse Foci

The foci of an ellipse are two special points that, along with any point on the ellipse, help satisfy a certain distance property. Their calculation involves using the values derived from the standard form of the ellipse equation. Generally expressed as \(c\), the foci relate to \(a\) and \(b\) through the equation \(c^2 = |a^2 - b^2|\).With our transformed equation, we can easily spot that:

• \(a\) is associated with the larger denominator, which corresponds to the direction of the major axis.
• Calculate \(c\) using the formula \(c^2 = |a^2 - b^2|\).
• From the center, move \(c\) units along the major axis in both directions to locate the foci.

This tells us exactly where to place the foci on the ellipse, providing essential geometric and algebraic insight into the structure and behavior of the ellipse.

Ellipse Vertices

Vertices of an ellipse are crucial points that lie at the end of both the major and minor axes, representing the furthest and closest points from the center along these axes. From the standard form of the ellipse equation, the vertices can be found easily. The process involves:

• Identifying \(a\) and \(b\) from the denominators in the standard form equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• Adding and subtracting \(a\) from the x-coordinate of the center to find horizontal vertices.
• Similarly, adding and subtracting \(b\) from the y-coordinate to find vertical vertices.

Essentially, the vertices not only define the maximum extent of the ellipse but also give us geometric cues about its overall shape and orientation on the plane.