Problem 12
Question
Graph each equation. Identify the conic section and describe the graph and its lines of symmetry. Then find the domain and range. $$ 4 x^{2}+4 y^{2}-20=0 $$
Step-by-Step Solution
Verified Answer
The conic section is a circle with center at the origin (0, 0) and a radius of \(\sqrt{5}\). Its lines of symmetry are the x-axis (y=0) and y-axis (x=0). The domain and range are both \([- \sqrt{5}, \sqrt{5}]\).
1Step 1: Identify Conic Section Type
The given equation is in the form \(Ax^{2} + By^{2} + C = 0\) where A = B = 4 and C = -20. Now, when A and B are the same, and both are not equal to zero, the equation represents a circle. Therefore, the given equation represents a circle.
2Step 2: Calculate Radius and Center
This is a circle equation in standard form which is \((x-h)^{2} + (y-k)^{2} = r^{2}\) where (h,k) is the center of the circle and r is its radius. We know that our equation is \(4 x^{2} +4 y^{2} = 20\). We can factor out 4 to get \(x^{2} + y^{2} = 5\). This tells us that our circle is centered at the origin (0,0) and has a radius of \(\sqrt{5}\).
3Step 3: Sketch the Graph and Identify Lines of Symmetry
Sketch a circle centered at the origin with a radius of \(\sqrt{5}\). The lines of symmetry for a circle are the x and y axes, assuming the center of the circle is at the origin. So for this circle, the lines of symmetry are x=0 and y=0, which are the y-axis and the x-axis, respectively.
4Step 4: Find Domain and Range
The domain and range for a circle equation are based on the radius and center of the circle. The domain of this function, which represents all possible x-values, is from -\(\sqrt{5}\) to \(\sqrt{5}\), or \([- \sqrt{5},\sqrt{5}]\). The range, representing all possible y-values, has the same interval as the domain, since the circle is centered at origin, which is \([- \sqrt{5}, \sqrt{5}]\).
Key Concepts
Graphing CirclesLines of SymmetryDomain and RangeIdentifying Equations of Shapes
Graphing Circles
Graphing a circle starts with identifying the standard form of the circle's equation, \[(x-h)^2 + (y-k)^2 = r^2\] where
Now, notice this equation matches the standard circle equation with the center at the origin (0,0) and a radius \(\sqrt{5}\).
To graph this circle, plot a point at the center and use the radius to draw the circle that extends out \(\sqrt{5}\) units in all directions. A precise graph provides a better understanding of how the circle behaves on the coordinate plane.
- (h, k) is the center of the circle
- r is the radius
Now, notice this equation matches the standard circle equation with the center at the origin (0,0) and a radius \(\sqrt{5}\).
To graph this circle, plot a point at the center and use the radius to draw the circle that extends out \(\sqrt{5}\) units in all directions. A precise graph provides a better understanding of how the circle behaves on the coordinate plane.
Lines of Symmetry
Symmetry in geometry refers to a shape looking identical when flipped or rotated around an axis.
For a circle like the one in this exercise, symmetry is remarkable.
Since our circle is centered at the origin, the x-axis and y-axis serve as the lines of symmetry. A circle centered at the origin means
For a circle like the one in this exercise, symmetry is remarkable.
Since our circle is centered at the origin, the x-axis and y-axis serve as the lines of symmetry. A circle centered at the origin means
- when cut along the x-axis, the top and bottom halves mirror each other
- when cut along the y-axis, the left and right halves mirror each other
Domain and Range
The domain and range indicate the extent of the circle on the x-axis and y-axis, respectively.
The domain outlines all possible x-values, while the range lists all possible y-values for points on the circle.
Given the equation \(x^2 + y^2 = 5\), we deduce the center of our circle is at the origin.
The radius of \(\sqrt{5}\) implies that:
The domain outlines all possible x-values, while the range lists all possible y-values for points on the circle.
Given the equation \(x^2 + y^2 = 5\), we deduce the center of our circle is at the origin.
The radius of \(\sqrt{5}\) implies that:
- The domain stretches from \(-\sqrt{5}\) to \(+\sqrt{5}\), written as \([-\sqrt{5}, \sqrt{5}]\).
- Similarly, the range also spans \([-\sqrt{5}, \sqrt{5}]\).
Identifying Equations of Shapes
Identifying conic sections from equations involves recognizing the form of the equation. For circles, equations often take the form \(Ax^2 + By^2 + C = 0\), where A = B.
In this example, both A and B are equal to 4, instantly revealing that this is a circle as both squared terms have equal coefficients. Understanding this form helps quickly categorize the shape:
In this example, both A and B are equal to 4, instantly revealing that this is a circle as both squared terms have equal coefficients. Understanding this form helps quickly categorize the shape:
- If A and B are the same and non-zero, you have a circle.
- If A and B are different, further analysis reveals ellipses, hyperbolas, or parabolas based on other terms and coefficients.
Other exercises in this chapter
Problem 12
Find the foci of each hyperbola. Then draw the graph. $$ \frac{x^{2}}{121}-\frac{y^{2}}{144}=1 $$
View solution Problem 12
Write an equation of a parabola with a vertex at the origin and the given focus. focus at \((0,-5)\)
View solution Problem 12
Write an equation for each translation. $$ x^{2}+y^{2}=81 ; \text { left } 1 \text { and } \operatorname{up} 3 $$
View solution Problem 13
Find an equation of an ellipse for each given height and width. Assume that the center of the ellipse is \((0,0) .\) $$ h=14 \text { yd, } w=28 $$
View solution