Problem 4 - 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. $$ x^{2}+4 y^{2}+2 x-24 y+33=0 $$

Step-by-Step Solution

Answer

The graph is an ellipse.

Step 1: Group and rearrange terms

Let's group the terms with respect to the variables: terms with \(x\) and terms with \(y\). The equation is: \[x^2 + 4y^2 + 2x - 24y + 33 = 0\] Group terms as follows: \[ (x^2 + 2x) + (4y^2 - 24y) = -33 \]

Step 2: Complete the square for \(x\) terms

Complete the square for the \(x\) terms \((x^2 + 2x)\).To complete the square, take half of the coefficient of \(x\), square it, and add and subtract it: \[ x^2 + 2x = (x^2 + 2x + 1 - 1) = (x + 1)^2 - 1 \]

Step 3: Complete the square for \(y\) terms

Complete the square for the \(y\) terms \((4y^2 - 24y)\). First factor out 4:\[ 4(y^2 - 6y) \]Take half of the coefficient of \(y\), square it, and add and subtract inside the parentheses:\[ y^2 - 6y = (y^2 - 6y + 9 - 9) = (y - 3)^2 - 9 \]Thus,\[ 4(y^2 - 6y) = 4((y - 3)^2 - 9) = 4(y - 3)^2 - 36 \]

Step 4: Substitute back into the equation

Now substitute the completed squares back into the equation:\[ (x + 1)^2 - 1 + 4(y - 3)^2 - 36 = -33 \] Simplify by combining like terms:\[ (x + 1)^2 + 4(y - 3)^2 = 4 \]

Step 5: Standard form and identify the graph type

Divide the entire equation by 4 to express in standard form:\[ \frac{(x + 1)^2}{4} + \frac{(y - 3)^2}{1} = 1 \]This is an equation of an ellipse because it has the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \(a^2\) and \(b^2\) are not equal.

Key Concepts

Standard Form

In mathematics, specifically in the study of algebraic curves, the term "standard form" refers to a preferred or simplified arrangement of an equation. The usefulness of placing equations in standard form is that it makes it much easier to identify the type of conic section being dealt with, such as a circle, ellipse, parabola, or hyperbola.
For ellipses, the standard form of the equation must be in the format:

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

Here, \((h, k)\) represents the center of the ellipse, while \(a^2\) and \(b^2\) represent the squares of the ellipse's radii in the horizontal and vertical directions, respectively.
Expressing an equation like \( (x+1)^2 + 4(y-3)^2 = 4 \) in standard form lets us clearly see it as an ellipse. By dividing by 4, each term becomes a part of the equation showing distinct ellipsoidal properties, critical for graphing and further analysis.

Ellipse

An ellipse is one of the conic sections formed by intersecting a plane with a double cone. Unlike circles, ellipses have two different axes of symmetry, known as the major and minor axes. The equation of an ellipse in standard form allows us to determine its shape and orientation.
When looking at the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), we can identify key features of the ellipse:

• Center at \((h, k)\), which is the midpoint of the major and minor axes.
• \(a\) is the semi-major axis length, and \(b\) is the semi-minor axis length.
• If \(a > b\), the major axis is horizontal; if \(b > a\), the major axis is vertical.

For the equation \( \frac{(x + 1)^2}{4} + \frac{(y - 3)^2}{1} = 1 \), such details show that the ellipse is centered at \((-1, 3)\), with a horizontal major axis and lengths being evident from \(a = 2\) and \(b = 1\). This breakdown helps in visualizing and graphing the ellipse successfully.

Completing the Square

To transform an equation into a standard form, particularly in conics, completing the square is often an essential step. This technique rearranges equations to highlight specific properties, such as centers or vertices.
Completing the square involves these basic steps:

• Take the quadratic expression, such as \(x^2 + bx\), and add and subtract the square of half the linear term's coefficient. For example, in \(x^2 + 2x\), half of 2 is 1, and squaring it gives 1. Thus, \((x^2 + 2x)\) becomes \((x + 1)^2 - 1\).
• This method applies similarly to \(y\) terms, but factor out coefficients from grouped terms first, like in the \(4(y^2 - 6y)\) example, which factors to \(4((y - 3)^2 - 9)\).

By using this method, terms are simplified into perfect squares, which are vital in expressing the equation as a standard form of an ellipse or other conic sections. It is a powerful operation for simplifying complex equations into understandable geometric shapes.