Problem 42 - College Algebra

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. $$ 64 x^{2}+128 x+9 y^{2}-72 y-368=0 $$

Step-by-Step Solution

Answer

The ellipse's center is (-1, 4), vertices are at (-1, 12) and (-1, -4), foci are at (-1, 4±√55).

Step 1: Rearrange and Group Terms

Firstly, we rearrange and group the terms of the given ellipse equation: \[64x^2 + 128x + 9y^2 - 72y = 368.\] Group the \(x\) terms and \(y\) terms together.

Step 2: Complete the Square for x-terms

Factor out the coefficient of \(x^2\) from the \(x\) terms:\[64(x^2 + 2x).\] Now, complete the square: \[x^2 + 2x = (x+1)^2 - 1.\] Therefore, \[64(x^2 + 2x) = 64((x+1)^2 - 1) = 64(x+1)^2 - 64.\]

Step 3: Complete the Square for y-terms

Factor out the coefficient of \(y^2\) from the \(y\) terms: \[9(y^2 - 8y).\]Complete the square: \[y^2 - 8y = (y-4)^2 - 16.\]Thus, \[9(y^2 - 8y) = 9((y-4)^2 - 16) = 9(y-4)^2 - 144.\]

Step 4: Simplify the Ellipse Equation

Substitute the results from completing the square into the equation and simplify:\[64(x+1)^2 - 64 + 9(y-4)^2 - 144 = 368.\]Combine like terms:\[64(x+1)^2 + 9(y-4)^2 = 576.\]

Step 5: Divide to Normalize the Ellipse Equation

Divide every term by 576 to bring it into standard form:\[\frac{64(x+1)^2}{576} + \frac{9(y-4)^2}{576} = 1.\]This simplifies to:\[\frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1.\]

Step 6: Identify Center, Vertices, and Foci of the Ellipse

The standard form of the ellipse equation is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where \((h, k)\) is the center.Here, the center is \((-1, 4)\), \(a^2 = 9\) and \(b^2 = 64\), so \(a = 3\) and \(b = 8\).Since \(b > a\), the major axis is vertical. Vertices are at \((-1, 4\pm 8)\) or \((-1, 12)\) and \((-1, -4)\).For the foci, use the formula \(c = \sqrt{b^2 - a^2}\):\(c = \sqrt{64 - 9} = \sqrt{55}.\)The foci are located at \((-1, 4\pm\sqrt{55})\).

Step 7: Graph the Ellipse

Plot the center at \((-1, 4)\). Mark the vertices at \((-1, 12)\) and \((-1, -4)\), and the co-vertices at \((2, 4)\) and \((-4, 4)\). Place the foci at \((-1, 4+\sqrt{55})\) and \((-1, 4-\sqrt{55})\). Draw an ellipse through these points.

Key Concepts

Foci Calculation

In an ellipse, the foci are two distinct points located along the major axis. To find the foci, we first need to identify the terms in the standard form of the ellipse: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \]where \((h, k)\) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. However, when \(b > a\), the roles are reversed, making the major axis vertical in this problem. To determine the distance of the foci from the center, we use the formula:\[ c = \sqrt{b^2 - a^2} \]where \(c\) represents that distance. Here, with \(a = 3\) and \(b = 8\), the calculation yields:\[ c = \sqrt{64 - 9} = \sqrt{55}. \]Thus, the foci are located at the points along the vertical axis:

• \((-1, 4+\sqrt{55})\)
• \((-1, 4-\sqrt{55})\)

These points are equidistant from the center and symmetrically positioned.

Completing the Square

Completing the square is a technique used to transform quadratic expressions into perfect squares, making it easier to work with equations involving polynomials. We apply this technique separately to both the \(x\) and \(y\) terms in the ellipse equation. Step-by-Step for the \(x\)-terms:

• Start with the expression: \(64(x^2 + 2x)\).
• Rewrite it by completing the square: \(x^2 + 2x = (x+1)^2 - 1\).
• After factoring out the 64: \[64(x^2 + 2x) = 64((x+1)^2 - 1) = 64(x+1)^2 - 64.\]

This turns the \(x\)-terms into a perfect square.Step-by-Step for the \(y\)-terms:

• Start with the expression: \(9(y^2 - 8y)\).
• Complete the square: \(y^2 - 8y = (y-4)^2 - 16\).
• After factoring out the 9: \[9(y^2 - 8y) = 9((y-4)^2 - 16) = 9(y-4)^2 - 144.\]

By completing the square, we reveal the terms related to the ellipse's center and axes.

Standard Form of Ellipse

The standard form of an ellipse is crucial for identifying its properties, such as its center, axes, and orientation. The general standard form equation for an ellipse is:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \]where \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.In our specific problem, after completing the square and simplifying, the ellipse equation becomes:\[ \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1. \]Here:

• \((x+1)^2\) shows the transformation required to discover the x-location of the center: \((-1)\).
• \((y-4)^2\) reveals the y-location of the center: \(4\).
• \(a^2 = 9\) gives \(a = 3\) and \(b^2 = 64\) gives \(b = 8\).

The ellipse's equation indicates vertical orientation since \(b > a\). This is key to sketching its shape correctly.

Center of Ellipse

Finding the center of an ellipse is especially straightforward once the equation is in standard form. The center is represented by the point \((h, k)\), which is derived from:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1. \]In this problem, upon completing the square for the x- and y-terms and arranging the equation into its standard form, we have:\[ \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1. \]From this equation, it is clear that:

• \(h = -1\)
• \(k = 4\)

Therefore, the center of the ellipse is at the coordinates \((-1, 4)\).The center serves as the origin point from which we measure the semi-major \((b)\) and semi-minor \((a)\) axes, as well as the distance to the foci.