Problem 60 - College Algebra with Modeling and Visualization

Question

Write the equation in standard form for an ellipse centered at (h, k). Identify the center and the vertices. $$ 16 x^{2}+48 x+4 y^{2}-20 y+57=0 $$

Step-by-Step Solution

Answer

Center: (-3/2, 5/2); Vertices: (-2, 5/2), (-1, 5/2), (-3/2, 3/2), (-3/2, 7/2).

Step 1: Rearrange the Equation

Start by rearranging the given equation to group the x terms and y terms: \[16x^2 + 48x + 4y^2 - 20y = -57\] This will prepare for completing the square.

Step 2: Factor and Complete the Square for X Terms

Factor out 16 from the x terms, yielding \[16(x^2 + 3x)\]. To complete the square, take half of the coefficient of x, square it, and add/subtract inside the parentheses: \[16(x^2 + 3x + \frac{9}{4} - \frac{9}{4})\] which simplifies to \[16((x + \frac{3}{2})^2 - \frac{9}{4})\]. Expand to \[16(x+\frac{3}{2})^2 - 36\].

Step 3: Factor and Complete the Square for Y Terms

Similarly, factor out 4 from the y terms: \[4(y^2 - 5y)\]. Complete the square by taking half of -5, squaring it, and adjusting: \[4(y^2 - 5y + 6.25 - 6.25)\], resulting in \[4((y - \frac{5}{2})^2 - 6.25)\]. Expand to \[4(y-\frac{5}{2})^2 - 25\].

Step 4: Substitute Back and Simplify Equation

Substitute back into the rearranged equation: \[16(x+\frac{3}{2})^2 - 36 + 4(y-\frac{5}{2})^2 - 25 = -57\].Combine all constant terms to get \[16(x+\frac{3}{2})^2 + 4(y-\frac{5}{2})^2 = 4\].

Step 5: Divide Through to Put Into Standard Form

Divide everything by 4 to conform with ellipse standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\): \[\frac{(x+\frac{3}{2})^2}{\frac{1}{4}} + \frac{(y-\frac{5}{2})^2}{1} = 1\].

Step 6: Identify Center and Vertices

The equation in standard form reveals a center at \((-\frac{3}{2}, \frac{5}{2})\).The lengths of the semi-axes are determined by \(a^2 = \frac{1}{4}, b^2 = 1\), thus \(a = \frac{1}{2}\) and \(b = 1\).The vertices along x-axis are at \((-2, \frac{5}{2})\) and \((-1, \frac{5}{2})\), while the vertices on the y-axis are at \((-\frac{3}{2}, \frac{3}{2})\) and \((-\frac{3}{2}, \frac{7}{2})\).

Key Concepts

Standard Form of an Ellipse

The standard form of an ellipse's equation is expressed as \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \((h, k)\) is the center of the ellipse, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. This form is essential to easily identify an ellipse's orientation and dimensions.
Depending on whether \(a > b\) or \(b > a\), the ellipse will stretch further along the x-axis or y-axis.

• If \(a > b\), the ellipse is elongated horizontally.
• If \(b > a\), the ellipse is elongated vertically.

This structure can appear daunting at first glance, but breaking it down into simpler components helps in identifying the critical elements of an ellipse.

Completing the Square

Completing the square is a vital technique used to transform a quadratic equation into a perfect square trinomial. It makes it simpler to convert a quadratic equation into the standard form of an ellipse.
To complete the square:

• First, take any quadratic term, for example, \(x^2 + bx\).
• Next, take half of the coefficient of \(x\), which is \(b\), square it, and add and subtract it inside the equation.

Let's say the term is \(16(x^2 + 3x)\). Take half of 3, square it to get \(\left(\frac{3}{2}\right)^2\) or \(\frac{9}{4}\), and adjust the equation: \(16(x^2 + 3x + \frac{9}{4} - \frac{9}{4})\). Factor it into \(16((x + \frac{3}{2})^2 - \frac{9}{4})\).
This alteration helps reveal the center and axis lengths needed for the standard form conversion.

Center of an Ellipse

In the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), \((h, k)\) is the center of the ellipse. It represents the point equidistant from all parts of the ellipse. To find the center after completing the square:

• Identify the part of the factored squared expression: \((x-h)\) gives \(h\) for the x-coordinate.
• The term \((y-k)\) gives \(k\) for the y-coordinate.

For example, from the given equation's standard form, the center is \((-\frac{3}{2}, \frac{5}{2})\). This point is crucial as it dictates where the ellipse is positioned on the graph.

Vertices of an Ellipse

Vertices are the most stretched-out points of an ellipse. With a center \((h, k)\), the vertices can be located by moving \(a\) units along the x or y-axis, depending on whether \(a > b\) or vice versa. For the equation \(\frac{(x+\frac{3}{2})^2}{\frac{1}{4}} + \frac{(y-\frac{5}{2})^2}{1} = 1\):

• From the center \((-\frac{3}{2}, \frac{5}{2})\), the x-axis vertices are \((-2, \frac{5}{2})\) and \((-1, \frac{5}{2})\) since \(a = \frac{1}{2}\).
• The y-axis vertices are \((-\frac{3}{2}, \frac{3}{2})\) and \((-\frac{3}{2}, \frac{7}{2})\) where \(b = 1\).

Understanding how far the vertices extend allows one to sketch and understand the ellipse's shape and location on the graph.