Problem 59 - 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. $$ 16 x^{2}-16 x+4 y^{2}+12 y=51 $$

Step-by-Step Solution

Answer

Center: \(\left(\frac{1}{2}, -\frac{3}{2}\right)\); Vertices: \((-\frac{3}{2}, -\frac{3}{2})\), \((\frac{5}{2}, -\frac{3}{2})\), \(\left(\frac{1}{2}, \frac{5}{2}\right)\), \(\left(\frac{1}{2}, -\frac{11}{2}\right)\).

Step 1: Organize Terms by Variable

Group the equation by variables: \[16x^2 - 16x + 4y^2 + 12y = 51\] becomes \[(16x^2 - 16x) + (4y^2 + 12y) = 51\].

Step 2: Factor Out Coefficients

Factor out the coefficients of the quadratic terms:\[16(x^2 - x) + 4(y^2 + 3y) = 51\].

Step 3: Complete the Square for x-terms

Complete the square for the expression \(x^2 - x\). Take half of the coefficient of \(x\), square it, and add inside the parenthesis:\[x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4}\].

Step 4: Substitute x-completed Square

Substitute back into the equation using the completed square:\[16((x - \frac{1}{2})^2 - \frac{1}{4})\]. Expand and simplify to:\[16(x - \frac{1}{2})^2 - 4\].

Step 5: Complete the Square for y-terms

Complete the square for the expression \(y^2 + 3y\). Take half of the coefficient of \(y\), square it, and add inside the parenthesis:\[y^2 + 3y = (y + \frac{3}{2})^2 - \frac{9}{4}\].

Step 6: Substitute y-completed Square

Substitute back into the equation using the completed square:\[4((y + \frac{3}{2})^2 - \frac{9}{4})\]. Expand and simplify to:\[4(y + \frac{3}{2})^2 - 9\].

Step 7: Simplify Entire Equation

Substitute completed square expressions and simplify entire equation:\[16(x - \frac{1}{2})^2 - 4 + 4(y + \frac{3}{2})^2 - 9 = 51\] which becomes:\[16(x - \frac{1}{2})^2 + 4(y + \frac{3}{2})^2 = 64\].

Step 8: Divide by Sum Constant

Divide the entire equation by 64 to bring to standard form:\[\frac{(x - \frac{1}{2})^2}{4} + \frac{(y + \frac{3}{2})^2}{16} = 1\].

Step 9: Identify Center and Vertices

The standard form of the ellipse is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]. Here, \(h = \frac{1}{2}\), \(k = -\frac{3}{2}\), \(a^2 = 4\) and \(b^2 = 16\).The center is \(\left(\frac{1}{2}, -\frac{3}{2}\right)\). Vertices along the x-axis are at \((\frac{1}{2} \pm 2, -\frac{3}{2})\) and along the y-axis at \((\frac{1}{2}, -\frac{3}{2} \pm 4)\).

Key Concepts

Standard form of an ellipse

The standard form for an ellipse's equation allows you to easily identify its center, axes length, and orientation. It is typically expressed as:

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

Here,

• \((h, k)\) represents the ellipse's center.
• \(a^2\) and \(b^2\) denote the squares of the semi-major and semi-minor axes, respectively.
• The form indicates what adjustments need to be made on the variables \(x\) and \(y\) to transform the basic circle equation \((x^2 + y^2 = 1)\).

If \(a > b\), the ellipse is elongated along the x-axis, making it a horizontal ellipse. Conversely, if \(b > a\), it's lengthier along the y-axis, indicating a vertical stretch. This formula is crucial as it provides an immediate visual understanding of the elliptical shape in relation to the coordinate plane.

Completing the square

Completing the square transforms a quadratic equation into a perfect square trinomial, making it easier to work with. This technique is particularly useful when dealing with conic sections like ellipses. Let's break this down:

• For an expression like \(x^2 - x\), to complete the square, take half of the coefficient of \(x\) (which is 1), square it to get \(\frac{1}{4}\), and add and subtract this inside the parenthesis: \(x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4}\).
• Similarly, for \(y^2 + 3y\), take half of 3, square it to get \(\frac{9}{4}\), and use it inside the expression: \(y^2 + 3y = (y + \frac{3}{2})^2 - \frac{9}{4}\).

This technique effectively transforms and simplifies equations, making it more manageable and allowing for easy conversion to the ellipse's standard form.

Ellipse center and vertices

Identifying the center and vertices is key to understanding the geometry of an ellipse. The center of an ellipse is at \((h, k)\), derived from the standard form equation:

• \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\).

In this format, \(h\) and \(k\) shift the position of the center from the origin. Once you identify the center, finding the vertices is simple:

• On the x-axis, vertices are \((h \pm a, k)\).
• On the y-axis, they are \((h, k \pm b)\).

In our example, once the expression is streamlined into the standard form, the ellipse is centered at \((\frac{1}{2}, -\frac{3}{2})\). The lengths of the semi-major and semi-minor axes (2 and 4, determined by \(a\) and \(b\) respectively) guide you to locate the vertices along both axes, thus painting a better picture of the ellipse's dimensions.