Problem 14 - Algebra 2

Question

Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center and the radius. For an ellipse or a hyperbola, give the center and the foci. Sketch the graph. \(3 x^{2}+6 x+y^{2}-6 y=-3\)

Step-by-Step Solution

Answer

The conic section represented by the given equation is an ellipse with its center at (-1,1), a horizontal semi-axis of 1, and a vertical semi-axis of \(sqrt(3)\). The ellipse does not have real foci, since their coordinates are imaginary.

Step 1: Rewrite the equation in standard form

Sort the given equation \(3 x^{2}+6 x+y^{2}-6 y=-3\) into the standard form. First, split the terms in x from those in y and rearrange the equation to get \(3x^2+6x + y^2-6y = -3\). Divide each term by 3 to simplify, getting \(x^2+2x + \frac{y^2}{3}-2y = -1\). Now, to complete the square for the x and y terms separately, take the coefficient of x, divide by 2 and square it. Add this square to both sides of the equation. Repeat the same step for the y-term.

Step 2: Identify the conic section and its characteristics

Completing the square gives \(x^2+2x+1 + \frac{(y-1)^2}{3}= 0\), which simplifies to \((x+1)^2+ \frac{(y-1)^2}{3}= 1\). This is an ellipse centered at (-1, 1) with a horizontal major axis. The horizontal semi-axis is 1 (the square root of the constant on the right) and the vertical semi-axis is \(sqrt(3)\). The foci can be found by using the formula \(c = sqrt(a^2 - b^2)\) where \(a = 1\) and \(b = sqrt(3)\), resulting in \(c = sqrt(1 - 3) = sqrt(-2) = i*sqrt(2)\). Since this is an imaginary number, there are no real foci.

Step 3: Sketch the graph

Plot the center at (-1, 1). Draw the major axis horizontally from the center with a length of 2 (2 times the horizontal semi-axis), and minor axis vertically with a length of 2sqrt(3) (2 times the vertical semi-axis). The ellipse is elongated vertically.

Key Concepts

Ellipse

An ellipse is a type of conic section that resembles an oval shape. In simple terms, it consists of all points for which the sum of the distances to two fixed points (called foci) is constant. The key features of an ellipse include:

• The center, which is the midpoint of the line segment joining the foci.
• The major axis, which is the longest diameter passing through the center.
• The minor axis, shorter than the major axis, also passing through the center.
• The foci, which are two fixed points on the major axis.

For this exercise, the conic section represented by the equation is an ellipse. The center of the ellipse is located at the point (-1, 1), and it has a horizontal major axis. As a helpful tip, remember that understanding the structure of an ellipse can offer insights into its geometric properties and aid in sketching its graph more precisely.

Standard Form

Many equations can be analyzed more easily if they're expressed in a particular way. The standard form of an ellipse is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where \((h, k)\) is the center of the ellipse, and \(a\) and \(b\) determine the lengths of the semi-major and semi-minor axes, respectively. If \(a > b\), the ellipse is horizontally stretched, and if \(b > a\), it is vertically stretched.In this exercise, the converted equation \((x+1)^2 + \frac{(y-1)^2}{3} = 1\) is in standard form. This form reveals much about the ellipse's geometry, such as the center \((-1, 1)\), a semi-major axis of length 1, and a semi-minor axis with length \(\sqrt{3}\)."

Completing the Square

Completing the square is a valuable technique aimed at converting quadratic terms into a perfect square trinomial. This process makes it easier to express certain mathematical expressions as squares, which is incredibly useful for converting equations into their standard forms.Here's a step-by-step approach to complete the square:

• Isolate the quadratic terms, typically in the form \(x^2 + bx\) or \(y^2 + cy\).
• Take the coefficient of the linear term (\(b\) or \(c\)), divide it by 2, and square it.
• Add and subtract this square inside the equation to maintain equality.

In the exercise, completing the square transformed \(x^2 + 2x\) into \((x+1)^2\) by adding and subtracting 1. Similarly, \(\frac{y^2}{3} - 2y\) transformed into \(\frac{(y-1)^2}{3}\), helping rearrange the equation into its standard form.

Equation Transformation

Equation transformation involves rewriting a given equation into a new format that can reveal important underlying structures. In the context of conic sections, transforming an equation helps in identifying the type of conic section and in understanding its essential characteristics.For the given equation, the strategy starts by reorganizing and simplifying it to become more manageable. It involves

• Identifying and separating terms involving different variables.
• Simplifying the equation by dividing by the necessary constants to ease calculation.
• Utilizing methods like completing the square to achieve a suitable form.

The equation \(3x^2 + 6x + y^2 - 6y = -3\) was transformed to \((x+1)^2 + \frac{(y-1)^2}{3} = 1\). This transformation allowed identifying it as an ellipse, providing a clear view of its center and axis dimensions. Mastering equation transformations is vital for diversifying problem-solving skills and approaches.