Problem 45 - College Algebra with Corequisite Support

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(4 x^{2}+16 x+4 y^{2}+16 y+16=0\)

Step-by-Step Solution

Answer

The ellipse is a circle centered at (-2, -2) with radius 2.

Step 1: Rearrange the Equation

Firstly, let's rearrange the given equation. We have: \[ 4x^2 + 16x + 4y^2 + 16y + 16 = 0 \] We notice that it is in standard quadratic form. Factor out the coefficient of 4 from both the \(x\) and \(y\) terms: \[ 4(x^2 + 4x) + 4(y^2 + 4y) = -16 \] Divide every term by 4 to simplify the equation: \[ (x^2 + 4x) + (y^2 + 4y) = -4 \]

Step 2: Complete the Square

Next, complete the square for both the \(x\) and \(y\) terms. For the \(x\) terms: \(x^2 + 4x\) becomes \((x+2)^2 - 4\). For the \(y\) terms: \(y^2 + 4y\) becomes \((y+2)^2 - 4\). Substitute these into the equation: \[(x + 2)^2 - 4 + (y + 2)^2 - 4 = -4 \] This simplifies to: \[ (x + 2)^2 + (y + 2)^2 = 4 \]

Step 3: Identify the Center and Axes

The equation \((x + 2)^2 + (y + 2)^2 = 4\) can be rewritten to form the standard form of an ellipse: \[ \frac{(x+2)^2}{4} + \frac{(y+2)^2}{4} = 1 \] This shows that the ellipse is centered at \((-2, -2)\) with both axes equal, indicating it is a circle. Since both denominators are 4, we find that the vertices are at \(-2 \pm 2, -2\) and \(-2, -2 \pm 2\). The foci are at the center of the circle since it is a unique case of an ellipse where the foci converge with the center.

Step 4: Sketch the Graph

To graph the ellipse (or circle, in this unique case), plot the center at \((-2, -2)\) on a coordinate plane. Since this is a circle with radius 2, plot points 2 units away from the center in all cardinal directions: \((-4, -2), (0, -2), (-2, 0),\) and \((-2, -4)\). Draw a smooth curve to form the circle, ensuring all plotted points are touched.

Key Concepts

Center of Ellipse

The center of an ellipse is a pivotal point around which the entire ellipse is symmetrically laid out. Finding the center helps in sketching and understanding the properties of the ellipse. In algebraic terms, an ellipse equation can be manipulated to reveal its center.

In the equation \[4x^2 + 16x + 4y^2 + 16y + 16 = 0,\] we rearranged terms and simplified as necessary to reveal the center coordinates. By completing the square for both the \(x\) and \(y\) variables, we derived the simplified equation:\[(x + 2)^2 + (y + 2)^2 = 4. \]This closely resembles the standard form of an ellipse. The expressions \((x+2)^2\) and \((y+2)^2\) show that the transformations involved include a horizontal shift of -2 and a vertical shift of -2. Therefore, the center of the ellipse is given by the point \((-2, -2)\).

Remember:

• The center is found at coordinates where \((x-h)^2\) and \((y-k)^2\) have been shifted.
• It acts as the midpoint between the vertices and foci of the ellipse.

Vertices of Ellipse

Vertices are crucial points on an ellipse that mark the extremities along its principal axes. In the standard form, the vertices are identified by measuring the distance from the center point.

From the standard form equation of the ellipse \[\frac{(x+2)^2}{4} + \frac{(y+2)^2}{4} = 1,\]the denominator under the \( (x+2)^2 \) and \( (y+2)^2 \) terms, which is 4, indicates that the semi-major or semi-minor axis length is 2, since \( \sqrt{4} = 2 \).

This being a circle (special case of an ellipse with equal semi-major and semi-minor axes), vertices are calculated by adding and subtracting 2 from the center's x-coordinate and y-coordinate.

• Horizontally at coordinates: \((-2 \, \pm \, 2, -2)\)
• Vertically at coordinates: \((-2, -2 \, \pm \, 2)\)

Thus, the vertices are \((-4, -2)\), \((0, -2)\), \((-2, 0)\), and \((-2, -4)\).

Vertices help in defining the overall shape and size of the ellipse, indicating how stretched or compact an ellipse is along its axes.

Foci of Ellipse

Foci are two fixed points inside an ellipse such that the sum of the distances from the foci to any point on the ellipse is constant. This unique property defines the general ellipse's geometry.

In the specific case of a circle, which our equation represents, the ellipse's foci coincide at the center point because of equal, balanced axes. Normally, the formula for the distance between the center and each focus point involves \[c = \, \sqrt{a^2 - b^2},\] where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. However, in this special circle ellipse, \[a = b,\] which leads to \[c = 0.\] Consequently, the foci both rest directly at the center, \((-2, -2)\).

Key takeaways include:

• In most ellipses, the foci are distinct points along the major axis.
• For circles, these points overlap at the center, illustrating differentiating characteristics.
• Foci are essential in applications like satellite dishes and optics, using their reflective properties.

Completing the Square

Completing the square is a valuable algebraic technique used to transform quadratic equations for easier analysis. It involves restructuring quadratic expressions into perfect squares.

For our given elliptical equation \[4x^2 + 16x + 4y^2 + 16y + 16 = 0,\] we needed this method to make the equation manageable and accessible for interpretation.

Steps to Complete the Square:

• Isolate Quadratic Terms: Factor out coefficients from \(x^2\) and \(y^2\) terms. Here, \(4\) was factored out.
• Balance the Equation: Shift constants to one side to set up for square completion.
• Form Perfect Squares: Add and subtract the necessary constant inside the parentheses to create squares. With expressions \(x^2 + 4x\) and \(y^2 + 4y\), the additions were 4.
• Normalize: Adjust the equation, ensuring the equation's balance remains by adjusting associated constants.

From this step-by-step approach emerged the compact, interpretable form \((x+2)^2 + (y+2)^2 = 4\), vital for readily identifying the ellipse's geometric elements.