Problem 78 - Intermediate Algebra

Question

Sketch the graph of each equation. If the graph is a parabola, find its vertex. If the graph is a circle, find its center and radius. $$\frac{x^{2}}{3}+\frac{y^{2}}{3}=2$$

Step-by-Step Solution

Answer

The graph is a circle with center (0, 0) and radius \(\sqrt{6}\).

Step 1: Identify the Type of Graph

The equation is given as \(\frac{x^{2}}{3} + \frac{y^{2}}{3} = 2\). This is similar to the general form of an ellipse equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). However, since 2 is on the right side, we initially simplify it to fit the standard form of an ellipse.

Step 2: Simplify the Equation to Standard Ellipse Form

Divide the entire equation by 2 to transform the equation into the standard form: \(\frac{x^{2}}{6} + \frac{y^{2}}{6} = 1\). This equation now represents an ellipse.

Step 3: Identify Key Parameters of the Ellipse

From \(\frac{x^{2}}{6} + \frac{y^{2}}{6} = 1\), recognize that this ellipse has equal values for the denominators, meaning it is a circle. The equation simplifies to \(x^{2} + y^{2} = 6\).

Step 4: Determine the Center and Radius of the Circle

Equation \(x^{2} + y^{2} = 6\) is in the form \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h, k)\) is the center and \(r\) is the radius. Here, \(h = 0\), \(k = 0\), and \(r^{2} = 6\), giving us \(r = \sqrt{6}\). Thus, the center is at (0, 0) and the radius is \(\sqrt{6}\).

Step 5: Sketch the Graph

Draw a circle with center at the origin \((0,0)\) in the coordinate plane and a radius of \(\sqrt{6}\). The circle should be a perfect round shape, passing through points approximately \(\pm\sqrt{6}\) units from the origin on both the x and y axes.

Key Concepts

Circle Equation

A circle equation is a specific kind of equation that represents all points in a plane equidistant from a given point, called the center. The standard form of a circle's equation is

• \((x-h)^2 + (y-k)^2 = r^2\)

Here,

• \((h,k)\) is the center of the circle, and
• \(r\) is the radius

The equation tells us that for any point

• \((x, y)\) on the circle
• Its distance to the center
• \((h, k)\)
• Is equal to the radius \(r\).

With a circle centered at the origin, the equation simplifies to

• \(x^2 + y^2 = r^2\).

This helps you quickly identify if an equation defines a circle and to understand its dimensions.

Center of Circle

The center of a circle is the fixed point around which the circle is perfectly symmetrical. In the generalized circle equation

• \((x-h)^2 + (y-k)^2 = r^2\),

the center of the circle is given by the point \((h, k)\). It acts as a reference point that is equidistant from all points on the perimeter of the circle. For equations where

• the center is at the origin,

\((h, k) = (0, 0)\),

• It means that the circle is centered precisely around the origin of the coordinate plane.

The position of the center influences the location of the whole circle, making it crucial for sketching and understanding the circle's place in a coordinate graph.

Radius of Circle

The radius of a circle is the distance from the center to any point on the circle's edge. It is a measure of how big the circle is. In the equation

• \((x-h)^2 + (y-k)^2 = r^2\),

\(r\) represents the radius. To find

• \(r\), take the square root of \(r^2\).

For example, given the circle equation

• \(x^2 + y^2 = 6\),

\(r^2 = 6\) so

• \(r = \sqrt{6}\).

This measurement helps define the size and scale of the circle on a graph. Having a longer radius makes for a larger circle, while a smaller radius means a more compact circle.

Standard Form of Ellipse

The standard form of an ellipse is an equation used to describe the shape and position of an ellipse on a coordinate plane. Typically written as

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

it generalizes the classic circle equation, allowing for unequal spreading across the x and y axes. Here,

• \(h\) and \(k\) denote the center,
• \(a\) and \(b\) are the semi-major and semi-minor axes.

When \(a\) is equal to \(b\),

• The shape is actually a circle.

Simplifying our original exercise's equation from

• \(\frac{x^2}{6} + \frac{y^2}{6} = 1\),

it was realized to be a circle with

• Equal denominators.

This form is pivotal in categorizing conic sections, not only showing when these sections are circular but also describing elongated or flattened ellipses.