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Magazine Chapter 11: Problem 24
Find the center, vertices, and foci of each ellipse and graph it. $$4 y^{2}+9 x^{2}=36$$
Short Answer
Expert verified
Center: (0,0). Vertices: (0, 3) and (0, -3). Foci: (0,鈭�5) and (0,-鈭�5).
Step by step solution
01
Rewrite the Equation in Standard Form
To identify the center, vertices, and foci, first rewrite the given equation in the standard form of an ellipse. The given equation is: \[ 4y^2 + 9x^2 = 36 \]. Divide both sides of the equation by 36 to isolate the ellipse form: \[ \frac{4y^2}{36} + \frac{9x^2}{36} = \frac{36}{36} \] which simplifies to \[ \frac{y^2}{9} + \frac{x^2}{4} = 1 \].
02
Identify the Center
The standard form of the ellipse \(\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1\) has its center at (0, 0) if there are no horizontal or vertical shifts. Here, \(\frac{y^2}{9} + \frac{x^2}{4} = 1\) directly implies that the center is at the origin (0, 0).
03
Determine the Lengths of the Axes
In the equation \(\frac{y^2}{9} + \frac{x^2}{4} = 1\), recognize the denominators describing the axis lengths. The larger denominator (9) is associated with the vertical axis. Thus, \(a^2 = 9\) which means \(a = 3\) and \(b^2 = 4\) which means \(b = 2\).
04
Find the Vertices
Since the larger denominator is under \(y^2\), this ellipse is vertical. The vertices are given by \(0, \pm a\) or (0, 卤3). So the vertices are at (0, 3) and (0, -3).
05
Compute the Foci
To find the foci, use the relationship \(c^2 = a^2 - b^2\). Here, \(a^2 = 9\) and \(b^2 = 4\), so: \[ c^2 = 9 - 4 = 5 \] which gives \[ c = \sqrt{5} \approx 2.24 \] , meaning the foci are at (0, 卤鈭�5).
06
Graph the Ellipse
Graphing the ellipse involves plotting the center (0,0), vertices (0,3) and (0,-3), and foci (0,鈭�5) and (0,-鈭�5). Draw an ellipse intersecting these points, ensuring the major axis is along the y-axis and the minor axis along the x-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ellipse Center
Locating the center of an ellipse is the foundational step in understanding its geometry. The given equation is \(4y^2 + 9x^2 = 36\). Start by rewriting it in the standard elliptical form \(\frac{y^2}{9} + \frac{x^2}{4} = 1 \). In this form, since there are no horizontal or vertical shifts, the center of the ellipse is at (0, 0). The center is important because it serves as the reference point from which other key features (like vertices and foci) are located.
Vertices of an Ellipse
To find the vertices of the ellipse, identify the lengths of the semi-major and semi-minor axes. The standard form \( \frac{y^2}{9} + \frac{x^2}{4} = 1 \) reveals these lengths through the denominators. The larger denominator (9) is under \(y^2\), indicating that the semi-major axis is along the y-axis, with \(a^2 = 9\), so \(a = 3\). Similarly, \(b^2 = 4\) gives \(b = 2\).
The vertices are located at the endpoints of the semi-major axis. Given the center at (0,0), the vertices of the ellipse are thus \( (0, \pm 3) \), which means they are at points (0, 3) and (0, -3). Vertices mark the farthest points of the ellipse along its longest dimension.
The vertices are located at the endpoints of the semi-major axis. Given the center at (0,0), the vertices of the ellipse are thus \( (0, \pm 3) \), which means they are at points (0, 3) and (0, -3). Vertices mark the farthest points of the ellipse along its longest dimension.
Foci of an Ellipse
The foci are two special points inside the ellipse that help define its shape. They are located along the major axis, closer to the center than the vertices. To find the foci, calculate \[ c^2 = a^2 - b^2 \]. For our ellipse here, \[ a^2 = 9 \] and \[ b^2 = 4 \], so \[ c^2 = 9 - 4 = 5 \] which gives \[ c = \sqrt{5} \].
Thus the foci are at \((0, \pm \sqrt{5})\), which is approximately \( (0, 卤2.24) \). The foci play a crucial role because the sum of the distances from these points to any point on the ellipse is constant.
Thus the foci are at \((0, \pm \sqrt{5})\), which is approximately \( (0, 卤2.24) \). The foci play a crucial role because the sum of the distances from these points to any point on the ellipse is constant.
Standard Form of an Ellipse
The standard form of an ellipse equation is crucial for identifying key features like its center, vertices, and foci. The given equation \(4y^2 + 9x^2 = 36\) must be rewritten to fit the standard form \(\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1\).
By dividing the entire equation by 36, it simplifies to \(\frac{y^2}{9} + \frac{x^2}{4} = 1 \). This transformation makes it easy to read off the relevant parameters: \( a^2 = 9 \) and \( b^2 = 4 \).
Once in this form, determining other properties of the ellipse becomes straightforward, including locating its center, defining the lengths of its axes, and calculating the positions of its foci.
By dividing the entire equation by 36, it simplifies to \(\frac{y^2}{9} + \frac{x^2}{4} = 1 \). This transformation makes it easy to read off the relevant parameters: \( a^2 = 9 \) and \( b^2 = 4 \).
Once in this form, determining other properties of the ellipse becomes straightforward, including locating its center, defining the lengths of its axes, and calculating the positions of its foci.
