Problem 12
Question
For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. \(\frac{x^{2}}{100}+\frac{y^{2}}{64}=1\)
Step-by-Step Solution
Verified Answer
Equation: \(\frac{x^2}{100} + \frac{y^2}{64} = 1\). Major axis endpoints: (10, 0), (-10, 0). Minor axis endpoints: (0, 8), (0, -8). Foci: (6, 0), (-6, 0).
1Step 1: Identify the Standard Form of the Ellipse
The given equation is \(\frac{x^2}{100} + \frac{y^2}{64} = 1\). This matches the standard form of an ellipse which is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.
2Step 2: Determine the Orientation of the Ellipse
Since \(100 > 64\), \(a^2 = 100\) and \(b^2 = 64\). This implies that \(a = 10\) and \(b = 8\). The ellipse has a horizontal major axis because \(a > b\).
3Step 3: Identify the Center of the Ellipse
The center of the ellipse in standard form is at the origin, which is \((0,0)\).
4Step 4: Find the Endpoints of the Major Axis
For a horizontal ellipse centered at \((0,0)\), the endpoints of the major axis are \((\pm a, 0)\). Therefore, the endpoints are \((10, 0)\) and \((-10, 0)\).
5Step 5: Find the Endpoints of the Minor Axis
For a horizontal ellipse centered at \((0,0)\), the endpoints of the minor axis are \((0, \pm b)\). Therefore, the endpoints are \((0, 8)\) and \((0, -8)\).
6Step 6: Calculate the Foci of the Ellipse
For an ellipse with a horizontal major axis, the foci are found at \((\pm c, 0)\), where \(c = \sqrt{a^2 - b^2} = \sqrt{100 - 64} = \sqrt{36} = 6\). Thus, the foci are located at \((6, 0)\) and \((-6, 0)\).
Key Concepts
Standard Form of EllipseMajor and Minor AxesFoci of Ellipse
Standard Form of Ellipse
When examining the equation of an ellipse, it's important to start by recognizing its standard form. An ellipse can be described by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-axes. This form is crucial because it lays the groundwork for understanding how the ellipse is oriented and where its axes lie.
In our exercise, the given equation \(\frac{x^2}{100} + \frac{y^2}{64} = 1\) exactly fits the standard form. We compare \(a^2\) and \(b^2\), and assign them to match \(a^2 = 100\) and \(b^2 = 64\). Thus, the values of \(a\) and \(b\) are 10 and 8, respectively. Recognizing this form immediately tells us the ellipse's fundamental properties such as its orientation and location.
In our exercise, the given equation \(\frac{x^2}{100} + \frac{y^2}{64} = 1\) exactly fits the standard form. We compare \(a^2\) and \(b^2\), and assign them to match \(a^2 = 100\) and \(b^2 = 64\). Thus, the values of \(a\) and \(b\) are 10 and 8, respectively. Recognizing this form immediately tells us the ellipse's fundamental properties such as its orientation and location.
- The larger denominator under the squared variable along the x-axis compared to the y-axis indicates a horizontal orientation.
- It is centered at the origin \((0, 0)\) as denoted by the absence of \((h, k)\) terms outside the squared variables.
Major and Minor Axes
Understanding the major and minor axes of an ellipse is integral for visualizing its shape and dimensions. The lengths of these axes are determined by the values \(a\) and \(b\), which we obtain from the standard form equation.
In our exercise, the major axis is horizontal because \(a = 10\) and that's larger than \(b = 8\). This means the ellipse stretches wider along the x-direction. Consequently, the endpoints of the major axis are calculated as \((\pm a, 0)\), giving us \((10, 0)\) and \((-10, 0)\).
The minor axis, being shorter, extends vertically, providing its endpoints as \((0, \pm b)\), which are \((0, 8)\) and \((0, -8)\).
In our exercise, the major axis is horizontal because \(a = 10\) and that's larger than \(b = 8\). This means the ellipse stretches wider along the x-direction. Consequently, the endpoints of the major axis are calculated as \((\pm a, 0)\), giving us \((10, 0)\) and \((-10, 0)\).
The minor axis, being shorter, extends vertically, providing its endpoints as \((0, \pm b)\), which are \((0, 8)\) and \((0, -8)\).
- The major axis defines the longest distance across the ellipse and passes through the center.
- The minor axis is the shortest width and is perpendicular to the major axis.
Foci of Ellipse
The foci of an ellipse are two significant points that lie along the major axis, helping to define its shape. They possess a unique property: the sum of the distances from each focus to any point on the ellipse is constant.
To find the foci of an ellipse, we use the formula \(c = \sqrt{a^2 - b^2}\). For our equation, this becomes \(c = \sqrt{100 - 64} = \sqrt{36} = 6\). Since the ellipse's major axis is horizontal, the foci are located symmetrically around the center along the x-axis, positioned at \((\pm c, 0)\), resulting in the coordinates \((6, 0)\) and \((-6, 0)\).
To find the foci of an ellipse, we use the formula \(c = \sqrt{a^2 - b^2}\). For our equation, this becomes \(c = \sqrt{100 - 64} = \sqrt{36} = 6\). Since the ellipse's major axis is horizontal, the foci are located symmetrically around the center along the x-axis, positioned at \((\pm c, 0)\), resulting in the coordinates \((6, 0)\) and \((-6, 0)\).
- Having the foci helps us understand how the ellipse is stretched along the major axis.
- They are pivotal in defining the precise curvature of the ellipse.
Other exercises in this chapter
Problem 12
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For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. \(r=\frac{4}{7+2 \quad \cos \theta}\)
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For the following exercises, determine which conic section is represented based on the given equation. \(3 x^{2}+6 x y+3 y^{2}-36 y-125=0\)
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