Problem 32
Question
Use a graphing device to graph the ellipse. $$ x^{2}+2 y^{2}=8 $$
Step-by-Step Solution
Verified Answer
The ellipse is centered at (0,0), extending about 2.83 units horizontally and 2 units vertically.
1Step 1: Write the Standard Form Equation
The general form of an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). To transform the given equation \( x^2 + 2y^2 = 8 \) to the standard form, divide every term by 8: \( \frac{x^2}{8} + \frac{y^2}{4} = 1 \). This shows that \( a^2 = 8 \) and \( b^2 = 4 \).
2Step 2: Identify Parameters of the Ellipse
From \( a^2 = 8 \), calculate \( a = \sqrt{8} = 2\sqrt{2} \). From \( b^2 = 4 \), calculate \( b = \sqrt{4} = 2 \). This defines the semi-major axis along the x-axis as \( a = 2\sqrt{2} \) and the semi-minor axis along the y-axis as \( b = 2 \).
3Step 3: Graph the Ellipse
Plot the ellipse using the parameters identified. Center the ellipse at the origin \((0,0)\). The ellipse stretches horizontally to \( \pm 2\sqrt{2} \) (approximately \( \pm 2.83 \)) and vertically to \( \pm 2 \). Use a graphing device to assist with drawing the precise shape.
Key Concepts
Standard Form Equation for EllipsesUnderstanding the Semi-Major AxisExplaining the Semi-Minor AxisGraphing Ellipses
Standard Form Equation for Ellipses
The standard form of an ellipse is crucial for understanding its geometry and graphing it accurately. An ellipse's standard form equation is given by:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]This equation allows us to identify the key parameters of the ellipse, specifically the semi-major and semi-minor axes. Notice that this form assumes the center of the ellipse is at the origin ((0,0)). In the exercise's equation, \( x^2 + 2y^2 = 8 \), there is a slight transformation required.
By dividing every term in the original equation by 8, we align it with the standard form:\[ \frac{x^2}{8} + \frac{y^2}{4} = 1 \]Now, the parameters \( a^2 \) and \( b^2 \) are visible: they are 8 and 4, respectively. Recognizing these values is essential as they determine the lengths of the axes and, thus, the exact shape of the ellipse.
By dividing every term in the original equation by 8, we align it with the standard form:\[ \frac{x^2}{8} + \frac{y^2}{4} = 1 \]Now, the parameters \( a^2 \) and \( b^2 \) are visible: they are 8 and 4, respectively. Recognizing these values is essential as they determine the lengths of the axes and, thus, the exact shape of the ellipse.
Understanding the Semi-Major Axis
The semi-major axis is the longest radius of an ellipse, running from its center to the furthest point on the perimeter. In our equation, it is associated with the larger of the two denominators found in the standard form.
After transforming the equation, we find that \( a^2 = 8 \). This leads us to calculate the semi-major axis, \( a \), as:\[ a = \sqrt{8} = 2\sqrt{2} \]This axis indicates that the ellipse stretches horizontally in its longest dimension. Knowing the semi-major axis is vital when graphing the ellipse, as it helps to determine the horizontal extent from the center, here located at \( ext{origin } (0,0) \).
After transforming the equation, we find that \( a^2 = 8 \). This leads us to calculate the semi-major axis, \( a \), as:\[ a = \sqrt{8} = 2\sqrt{2} \]This axis indicates that the ellipse stretches horizontally in its longest dimension. Knowing the semi-major axis is vital when graphing the ellipse, as it helps to determine the horizontal extent from the center, here located at \( ext{origin } (0,0) \).
- The ellipse will extend from the origin to both \( +2\sqrt{2} \) and \(-2\sqrt{2} \) along the x-axis.
- This translates approximately to the ellipse reaching \( ext{almost } \pm 2.83 \).
Explaining the Semi-Minor Axis
While the semi-major axis is the longest stretch of an ellipse, the semi-minor axis is the shortest. It grows from center to the edge perpendicularly to the major axis. In the transformed standard form, this shorter radius corresponds to the smaller denominator. In this situation, we find:
\( b^2 = 4 \)This gives us a semi-minor axis, \( b \), of:\[ b = \sqrt{4} = 2 \]
\( b^2 = 4 \)This gives us a semi-minor axis, \( b \), of:\[ b = \sqrt{4} = 2 \]
- The ellipse will extend vertically from the origin \( ext{upward at } +2 \).
- And also \( ext{downward to } -2 \) along the y-axis.
Graphing Ellipses
Graphing an ellipse might seem complex, but understanding its parameters simplifies the task. Once the axes are identified, the sketch becomes much clearer. For the ellipse given by\( \frac{x^2}{8} + \frac{y^2}{4} = 1 \):
- Center the ellipse at the origin ((0,0)).
- Mark the horizontal reach at \( +2\sqrt{2} \) and \(-2\sqrt{2} \).
- Mark the vertical reach at \( +2 \) and \(-2 \).
Other exercises in this chapter
Problem 32
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\(29-40\) Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Focus: \(F(5,0)\)
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Find an equation for the hyperbola that satisfies the given conditions. Foci: \((0, \pm 2),\) vertices: \((0, \pm 1)\)
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