Problem 31
Question
Graph equation. \(8 x^{2}+2 y^{2}=32\)
Step-by-Step Solution
Verified Answer
The equation represents a vertically oriented ellipse centered at (0,0) with axes lengths of 4 and 2.
1Step 1: Identify the Type of Conic Section
The given equation is \(8x^{2} + 2y^{2} = 32\). Notice that it is written in the standard form for conic sections where both \(x^{2}\) and \(y^{2}\) terms have positive coefficients. This indicates that the conic section is an ellipse.
2Step 2: Rewrite the Equation
To make identification easier, divide every term by 32 to simplify the equation: \(\frac{8x^{2}}{32} + \frac{2y^{2}}{32} = 1 \). Simplifying, we get \(\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1\).
3Step 3: Identify the Parameters of the Ellipse
The equation now is \(\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1\), which is in the standard form of \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), where \(a^{2} = 4\) and \(b^{2} = 16\). Thus, \(a = 2\) and \(b = 4\). This tells us the semi-major axis length is 4, and the semi-minor axis length is 2.
4Step 4: Determine Orientation and Position
Since \(b > a\), the ellipse's major axis is along the y-axis. It is centered at the origin (0,0) because there are no linear terms for \(x\) or \(y\), indicating no horizontal or vertical shifts.
5Step 5: Sketch the Ellipse
Plot the center of the ellipse at (0,0). The endpoints of the major axis will be at (0,4) and (0,-4), while the endpoints of the minor axis will be at (2,0) and (-2,0). The ellipse is elongated vertically due to its orientation.
Key Concepts
EllipseGraphing EquationsStandard Form
Ellipse
An ellipse is a type of conic section that appears as a stretched circle. It is defined by two main axes: the major axis and the minor axis. These axes cross at the center of the ellipse, which forms the ellipse's symmetry line. The major axis is the longest line through the center, while the minor axis is the shortest. The endpoints of these axes are important as they help to define the shape and orientation of an ellipse.
- The length from the center to the endpoint of the major axis is the semi-major axis.
- The length from the center to the endpoint of the minor axis is the semi-minor axis.
Graphing Equations
Graphing equations involves plotting a shape or curve on a coordinate plane using a function or equation. For conic sections like ellipses, this means identifying key parameters from the equation that dictate size, shape, and position.
Graphically representing an ellipse requires:
Graphically representing an ellipse requires:
- Identifying the center of the ellipse.
- Determining the lengths and orientation of the major and minor axes.
- Using the endpoints derived from the axes lengths to define the boundary of the ellipse.
Standard Form
The standard form of an ellipse's equation is crucial in quickly identifying essential characteristics of the ellipse. For ellipses centered at the origin, the standard form is \[\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\]where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.
This form helps by:
This form helps by:
- Allowing immediate identification of the lengths of the semi-major and semi-minor axes through \(a\) and \(b\).
- Providing information about the orientation of the ellipse, denoted by which value is larger, \(a\) or \(b\).
Other exercises in this chapter
Problem 30
Graph equation. \(25(x+3)^{2}+4(y-3)^{2}=100\)
View solution Problem 31
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=\sqrt{x}} \\ {x^{2}+y^{2}=12} \end{array}\right. $$
View solution Problem 31
The graph of each equation is a circle. Find the center and the radius and then graph the circle. $$x^{2}+y^{2}+2 x-4 y=4$$
View solution Problem 32
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=\sqrt{x}} \\ {x^{2}+y^{2}=20} \end{array}\right. $$
View solution