Chapter 13: Problem 2
Graph each ellipse. $$ \frac{x^{2}}{16}+\frac{y^{2}}{9}=1 $$
Short Answer
Expert verified
The ellipse is centered at (0,0) with a semi-major axis of 4 along the x-axis and a semi-minor axis of 3 along the y-axis.
Step by step solution
01
Identify the standard form of the ellipse
The given equation is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \). This matches the standard form of an ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a^2 = 16 \) and \( b^2 = 9 \).
02
Determine the lengths of the semi-major and semi-minor axes
From the equation, \( a^2 = 16 \) and \( b^2 = 9 \). Calculate \( a \) and \( b \) by taking the square roots: \( a = \sqrt{16} = 4 \) and \( b = \sqrt{9} = 3 \). Since \( a > b \), the semi-major axis is along the x-axis.
03
Find the center of the ellipse
The center of the ellipse from the standard form equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is at the origin (0,0) because there are no \( (x-h) \) or \( (y-k) \) shifts in the equation.
04
Sketch the ellipse
Draw the coordinate axes and plot the center at (0,0). Count 4 units along the x-axis to mark the vertices at (4,0) and (-4,0). Count 3 units along the y-axis to mark the co-vertices at (0,3) and (0,-3). Connect these points smoothly to form the ellipse.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Semi-Major Axis
The semi-major axis is an essential part of an ellipse, representing the longest radius from the center to the edge. For any standard ellipse, there are always two axes: the major axis and the minor axis. The semi-major axis is half of the major axis, and it determines the widest part of the ellipse.
In the given equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we identified that \( a^2 = 16 \). By taking the square root, we find that \( a = \sqrt{16} = 4 \). Thus, the semi-major axis measures 4 units from the center, along the x-axis.
When sketching an ellipse, you'll want to mark the endpoints of the semi-major axis, which determine its stretch horizontally or vertically, depending on which variable has the larger denominator in the given equation.
In the given equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we identified that \( a^2 = 16 \). By taking the square root, we find that \( a = \sqrt{16} = 4 \). Thus, the semi-major axis measures 4 units from the center, along the x-axis.
When sketching an ellipse, you'll want to mark the endpoints of the semi-major axis, which determine its stretch horizontally or vertically, depending on which variable has the larger denominator in the given equation.
Semi-Minor Axis
Just as the semi-major axis defines the stretch of an ellipse along its longest direction, the semi-minor axis provides the width in its shortest direction. In our standard form, the semi-minor axis is half of the minor axis of the ellipse.
From our equation, \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we found \( b^2 = 9 \), and by calculating \( b = \sqrt{9} = 3 \), we discover that the semi-minor axis is 3 units long.
The semi-minor axis is perpendicular to the semi-major axis, and since the semi-major is along the x-axis, here the semi-minor is aligned with the y-axis. Plotting the ellipse, you would see the shorter curve stretching 3 units above and below the center along the y-axis.
From our equation, \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we found \( b^2 = 9 \), and by calculating \( b = \sqrt{9} = 3 \), we discover that the semi-minor axis is 3 units long.
The semi-minor axis is perpendicular to the semi-major axis, and since the semi-major is along the x-axis, here the semi-minor is aligned with the y-axis. Plotting the ellipse, you would see the shorter curve stretching 3 units above and below the center along the y-axis.
Center of the Ellipse
At the heart of the ellipse is its center, a point equidistant from all points on the geometric figure. In standard equations of the form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), (h, k) denotes the center of the ellipse.
Since our given ellipse equation is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we can deduce that there are no horizontal or vertical shifts, implying that h = 0 and k = 0. Thus, the center of the ellipse is at the origin (0,0).
Understanding the center is vital, as it helps in plotting the ellipse accurately on a coordinate plane by using it as the pivot point for marking the axes and ensuring symmetrical curvature.
Since our given ellipse equation is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), we can deduce that there are no horizontal or vertical shifts, implying that h = 0 and k = 0. Thus, the center of the ellipse is at the origin (0,0).
Understanding the center is vital, as it helps in plotting the ellipse accurately on a coordinate plane by using it as the pivot point for marking the axes and ensuring symmetrical curvature.
Standard Form Equation
The standard form equation for an ellipse is a structured representation that helps us quickly determine several properties of the ellipse. It is expressed as \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) where (h, k) is the ellipse's center, and these fractions describe the ellipse's extent along each axis.
For our ellipse equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), the standard form reveals that the center is at (0,0), \( a^2 = 16 \), and \( b^2 = 9 \). This information allows us to derive the lengths of the semi-major axis (4 units) and the semi-minor axis (3 units).
The standard form is crucial because it provides a uniform way of conveying the shape and position of the ellipse on the coordinate grid, aiding in visualization and graphing.
For our ellipse equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), the standard form reveals that the center is at (0,0), \( a^2 = 16 \), and \( b^2 = 9 \). This information allows us to derive the lengths of the semi-major axis (4 units) and the semi-minor axis (3 units).
The standard form is crucial because it provides a uniform way of conveying the shape and position of the ellipse on the coordinate grid, aiding in visualization and graphing.