Problem 45 - A Graphical Approach to College Algebra

Question

Write the equation in standard form for an ellipse centered at ( \(h, k\) ). Identify the center and vertices. $$16 x^{2}-16 x+4 y^{2}+12 y=51$$

Step-by-Step Solution

Answer

Center: \((\frac{1}{2}, -\frac{3}{2})\); Vertices: \((2.5, -\frac{3}{2})\), \((-1.5, -\frac{3}{2})\), \((\frac{1}{2}, \frac{5}{2})\), \((\frac{1}{2}, -\frac{11}{2})\)

Step 1: Group and Rearrange the Terms

Group the terms with the same variable together and move the constant to the other side of the equation. You get: \[ 16(x^2 - x) + 4(y^2 + 3y) = 51 \]

Step 2: Complete the Square for x-terms

Factor out the coefficient of the quadratic term and complete the square for \(x\): \[ 16 \left( x^2 - x + \left(-\frac{1}{2}\right)^2 \right) - 16 \left( -\frac{1}{2} \right)^2 + 4y^2 + 12y = 51 \]This becomes:\[ 16(x - \frac{1}{2})^2 + 4y^2 + 12y = 51 + 16 \cdot \frac{1}{4} \]

Step 3: Complete the Square for y-terms

Next, complete the square for the \(y\) terms:\[ 4(y^2 + 3y + \left(\frac{3}{2}\right)^2) - 4 \left(\frac{3}{2}\right)^2 = 55 \]It simplifies to:\[ 4(y + \frac{3}{2})^2 = 55 + 4 \cdot \frac{9}{4} \]

Step 4: Simplify and Set in Standard Form

Combine the complete squares and simplify:\[ 16(x - \frac{1}{2})^2 + 4(y + \frac{3}{2})^2 = 64 \]Divide the entire equation by 64 to get the standard form of the ellipse equation:\[ \left(\frac{x - \frac{1}{2}}{2}\right)^2 + \left(\frac{y + \frac{3}{2}}{4}\right)^2 = 1 \]

Step 5: Identify Center and Vertices

From the standard form, identify the center of the ellipse at \( (h, k) = \left( \frac{1}{2}, -\frac{3}{2} \right) \).The vertices along the x-axis direction (major axis) are at \( (\frac{1}{2} \pm 2, -\frac{3}{2}) \), so they are: \((2.5, -\frac{3}{2})\) and \((-1.5, -\frac{3}{2})\).The vertices along the y-axis direction are at \( (\frac{1}{2}, -\frac{3}{2} \pm 4) \), so they are: \((\frac{1}{2}, \frac{5}{2})\) and \((\frac{1}{2}, -\frac{11}{2})\).

Key Concepts

Completing the Square

Completing the square is a method used in algebra to transform a quadratic expression into a perfect square trinomial. This is a key step in converting quadratic equations into a more manageable form. In the context of conic sections, it's essential for rewriting the equation of an ellipse into its standard form.

When faced with a quadratic term like \(16(x^2 - x)\), start by factoring out the coefficient of the \(x^2\) term, which in this case is 16. This simplifies our expression:

• Factor: \(16(x^2 - x)\ = 16(x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2)\)
• Add and subtract \(\left(-\frac{1}{2}\right)^2\) to complete the square within the brackets.
• The expression becomes a perfect square trinomial: \((x - \frac{1}{2})^2\)

Now, repeat similar steps for the \(y\) terms. With \(4(y^2 + 3y)\), factor out the 4 and adjust the expression:

• Factor: \(4(y^2 + 3y) = 4(y^2 + 3y + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2)\)
• Add and subtract \(\left(\frac{3}{2}\right)^2\) to complete the square.
• Now it becomes \((y + \frac{3}{2})^2\)

Completing the square is crucial for transforming the equation of an ellipse into its recognizable standard form, facilitating identification of important geometric properties.

Standard Form of Conic Sections

The standard form of conic sections helps to easily identify the various parts of the shapes such as ellipses, hyperbolas, and parabolas.

For an ellipse, the standard equation looks like this:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]This format reveals that:

• \((h, k)\) is the center of the ellipse.
• \(a\) and \(b\) represent the distances from the center to the vertices along the x-axis and y-axis respectively.

In our problem, through completing the square, we reach:\[16(x-\frac{1}{2})^2 + 4(y+\frac{3}{2})^2 = 64\]By dividing each term by 64, we convert it to:\[\left(\frac{x - \frac{1}{2}}{2}\right)^2 + \left(\frac{y + \frac{3}{2}}{4}\right)^2 = 1\]Here we recognize \(a = 2\) and \(b = 4\), corresponding to the horizontal and vertical distances from the center to the vertices. This standard form elucidates the shape and dimensions of the ellipse.

Vertices of an Ellipse

The vertices of an ellipse are critical points that define its size and alignment.

From the standard form equation \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify \(a\) and \(b\) as the semi-major and semi-minor axes, impacting the range and stretch of the ellipse.

The center of the ellipse \( (h, k) \) is \(\left( \frac{1}{2}, -\frac{3}{2} \right)\). The vertices along the ellipse’s major axis (larger value, \(b = 4\)) are located vertically due to it being related to the \(y\) term:

• Upper vertex: \(\left(\frac{1}{2}, -\frac{3}{2} + 4 \right) = \left(\frac{1}{2}, \frac{5}{2}\right)\)
• Lower vertex: \(\left(\frac{1}{2}, -\frac{3}{2} - 4 \right) = \left(\frac{1}{2}, -\frac{11}{2}\right)\)

The vertices along the ellipse’s minor axis (shorter value, \(a = 2\)) are horizontal:

• Right vertex: \(\left(\frac{1}{2} + 2, -\frac{3}{2} \right) = (2.5, -\frac{3}{2})\)
• Left vertex: \(\left(\frac{1}{2} - 2, -\frac{3}{2} \right) = (-1.5, -\frac{3}{2})\)

These vertices give a clear picture of the ellipse's boundaries and how it sits in relation to its center point.