Problem 5
Question
What can be said about the symmetry of the graph of an ellipse with center at the origin and foci along the \(y\) -axis?
Step-by-Step Solution
Verified Answer
The ellipse is symmetric about the x-axis, y-axis, and origin.
1Step 1: Understand the Ellipse Equation
An ellipse centered at the origin with foci along the \(y\)-axis has the standard form equation: \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \] where \(a > b\). Here, \(a\) is the semi-major axis length along the \(y\)-axis, and \(b\) is the semi-minor axis length along the \(x\)-axis.
2Step 2: Define Symmetry in the Context of an Ellipse
Symmetry in the context of a graph indicates that the graph looks the same when viewed from different perspectives or through various transformations such as reflection across axes or rotation around a point.
3Step 3: Check for Symmetry about the Y-Axis
Reflecting the ellipse equation \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) across the \(y\)-axis involves replacing \(x\) with \(-x\). After substitution, we get the same equation \( \frac{(-x)^2}{b^2} + \frac{y^2}{a^2} = 1 \), which is identical to the original equation. Thus, the ellipse is symmetric with respect to the \(y\)-axis.
4Step 4: Check for Symmetry about the X-Axis
Reflecting the ellipse equation across the \(x\)-axis involves replacing \(y\) with \(-y\). After substitution, the equation \( \frac{x^2}{b^2} + \frac{(-y)^2}{a^2} = 1 \) is obtained, which is exactly the same as the original equation. Therefore, the ellipse is symmetric with respect to the \(x\)-axis as well.
5Step 5: Check for Symmetry about the Origin
To check symmetry about the origin, replace \(x\) with \(-x\) and \(y\) with \(-y\), resulting in the equation \( \frac{(-x)^2}{b^2} + \frac{(-y)^2}{a^2} = 1 \). This is the same as the original equation, confirming that the ellipse is symmetric with respect to the origin.
Key Concepts
Ellipse EquationSymmetry in GraphsReflection Across AxesOrigin Symmetry
Ellipse Equation
An ellipse is a type of conic section that often suggests a stretched circle. When centered at the origin, its equation takes on a specific form written as: \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]Here, \(a\) and \(b\) are crucial as they represent the lengths of the semi-major and semi-minor axes of the ellipse. Notably, \(a\) must be greater than \(b\) if the foci are along the \(y\)-axis, indicating that the ellipse is taller than it is wide.
- \(x^2\) and \(y^2\) terms signify that the curve is centered at the origin.
- The coefficients \(b^2\) and \(a^2\) determine the spread of the ellipse along the x-axis and the y-axis respectively.
Symmetry in Graphs
Symmetry in graphs reveals how shapes or objects can look consistent from different viewpoints or after certain geometric transformations. In essence, this means:
- A graph is symmetric if it mirrors across a specific line, like an axis.
- It can also be symmetric if it rotates upon a point and looks unchanged.
Reflection Across Axes
Reflecting a function or equation across an axis is a powerful tool to determine symmetry. Here's how it applies to ellipses:
- Reflection across the \(y\)-axis: Change \(x\) to \(-x\). If the resulting equation is the same, it confirms \(y\)-axis symmetry.
- Reflection across the \(x\)-axis: Change \(y\) to \(-y\). If the equation remains unchanged, this indicates symmetry about the \(x\)-axis.
Origin Symmetry
Origin symmetry implies that an entire graph remains unchanged even after being flipped in both horizontal and vertical directions. Specifically, this involves replacing \(x\) with \(-x\) and \(y\) with \(-y\). For an ellipse equation:\[\frac{(-x)^2}{b^2} + \frac{(-y)^2}{a^2} = 1\]Since this version is identical to the original equation, the ellipse must be origin symmetric.
- Origin symmetry reveals that all points are equidistant along axes relative to the origin.
- This type of symmetry contributes to the holistic, balanced appearance of an ellipse.
Other exercises in this chapter
Problem 5
As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?
View solution Problem 5
Where must the center of hyperbola be relative to its foci?
View solution Problem 6
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. \(r=\frac{6}{1-2 \cos \theta}\)
View solution Problem 6
For the following exercises, determine which conic section is represented based on the given equation. \(9 x^{2}+4 y^{2}+72 x+36 y-500=0\)
View solution