Problem 37 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ x^{2}+4 y^{2}-11=2(4 y-x) $$

Step-by-Step Solution

Answer

The graph is an ellipse in standard form: \(\frac{(x+1)^2}{28} + \frac{(y-1)^2}{7} = 1\).

Step 1: Expand and Simplify the Equation

First, expand the right side of the equation: \[ 2(4y - x) = 8y - 2x \]Now, the equation becomes:\[ x^2 + 4y^2 - 11 = 8y - 2x \]Next, add \(2x\) and subtract \(8y\) from both sides to bring all terms to one side:\[ x^2 + 2x + 4y^2 - 8y - 11 = 0 \]

Step 2: Complete the Square for x Terms

To complete the square for \(x\), look at the terms involving \(x\):\[ x^2 + 2x \]Take half of the coefficient of \(x\), square it, and add it inside the equation:\[ x^2 + 2x + 1 = (x+1)^2 \]Since we added 1, also subtract 1 from the equation to maintain balance:\[ (x+1)^2 + 4y^2 - 8y - 12 = 0 \]

Step 3: Complete the Square for y Terms

Next, complete the square for \(y\):\[ 4y^2 - 8y \]Factor out the 4:\[ 4(y^2 - 2y) \]Take half of the coefficient of \(y\), square it, and add it inside the parenthesis:\[ 4(y^2 - 2y + 1) = 4(y-1)^2 \]Since we introduced \(4\), also subtract \(4\) to balance the equation:\[ (x+1)^2 + 4(y-1)^2 - 16 - 12 = 0 \]

Step 4: Simplify into Standard Form

Combine all constant terms:\[ (x+1)^2 + 4(y-1)^2 = 16 + 12 \]\[ (x+1)^2 + 4(y-1)^2 = 28 \]Divide the entire equation by 28 to express it in standard form:\[ \frac{(x+1)^2}{28} + \frac{(y-1)^2}{7} = 1 \]This is the standard form of an ellipse.

Step 5: Determine the Type of Conic Section

By analyzing the standard form, \(\frac{(x+1)^2}{28} + \frac{(y-1)^2}{7} = 1\), we see that it resembles the general form of an ellipse,\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] with different denominators for \(x\) and \(y\) terms, confirming it is an ellipse.

Step 6: Graph the Equation

The equation is an ellipse centered at \((-1, 1)\). The semi-major axis corresponds to \(\sqrt{28}\) along the \(x\)-axis (horizontal direction) and \(\sqrt{7}\) along the \(y\)-axis (vertical direction). Graph the ellipse by plotting these dimensions around the center \((-1, 1)\).

Key Concepts

Standard Form

The standard form for conic sections is an essential concept in understanding how different equations describe geometric shapes like ellipses, circles, parabolas, and hyperbolas. The standard form of an equation helps identify these shapes and transform them for easier graphing. For ellipses, the standard form is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where

• \((h, k)\) is the center of the ellipse.
• \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively.

To convert an equation to this standard form, we complete the square for both \(x\) and \(y\) terms, which allows us to group the squared terms neatly and identify essential parameters like the center and radii.

Ellipse

An ellipse is a fascinating conic section characterized by its oval shape. It can be thought of as a stretched circle, with two axes of symmetry: the major and minor axes. In an ellipse's standard equation, the denominators determine which axis is longer. The ellipse equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]has important features:

• If \(a > b\), the major axis is horizontal.
• If \(b > a\), the major axis is vertical.
• The points \((h \pm a, k)\) and \((h, k \pm b)\) are the vertices of the ellipse.

Identifying these parameters from an equation helps in understanding the ellipse's orientation and dimensions. The center \((h, k)\) is the midpoint of the ellipse, and the vertices mark the furthest extent along each axis, defining the ellipse's boundaries.

Completing the Square

Completing the square is a powerful algebraic technique used in transforming quadratic equations. It is especially useful for conic sections, where equations need to be adjusted to standard forms. The process involves:

• Grouping terms of the same variable together.
• Taking the coefficient of the linear term, halving it, and squaring the result.
• Adding and subtracting this square inside the equation to keep balance.

For example, for terms like \(x^2 + 2x\), we take half the coefficient of \(x\), which is 1, square it to get 1, and use it to complete the square: \[(x+1)^2\].
Similarly, with \(y^2 - 2y\), we'd square \(-1\) to complete the square to \[(y-1)^2\].
Completing the square enables us to reorganize the equation into a recognizable form, facilitating further analysis and graphing of the ellipse.

Graphing Ellipses

Graphing an ellipse involves interpreting its standard form and translating the mathematical equation into a visual representation. After converting the equation into standard form, several steps are taken:

• Identify the center \((h, k)\) of the ellipse.
• Determine the lengths of the semi-major and semi-minor axes, \(a\) and \(b\).
• Use \((h \pm a, k)\) and \((h, k \pm b)\) to locate the vertices along the major and minor axes.
• Draw a smooth oval passing through the four vertices, checking symmetry around both axes.

For example, an ellipse centered at \((-1, 1)\) with axes \(\sqrt{28}\) (horizontal) and \(\sqrt{7}\) (vertical) would be drawn by plotting these dimensions symmetrically around the center.
Graphing provides a complete picture of the ellipse, helping visualize its spread and position in the coordinate plane.