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Chapter 10: Problem 10

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$\frac{(x+2)^{2}}{25}+\frac{(y-3)^{2}}{4}=1$$

Short Answer

Expert verified
Vertices: (3, 3) and (-7, 3); Foci: (-2卤鈭�21, 3).

Step by step solution

01

Identify the Equation Type

The given equation is \(\frac{(x+2)^2}{25}+\frac{(y-3)^2}{4}=1\), which is the standard form of an ellipse centered at \((-2, 3)\). The large denominator under \((x+2)^2\), which is 25, indicates that the major axis is horizontal.
02

Determine the Lengths of the Axes

For an ellipse of the form \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), where \(a^2\) is the larger denominator, the major axis is along the x-axis. Here, \(a^2 = 25\) and \(b^2 = 4\), so \(a = \sqrt{25} = 5\) and \(b = \sqrt{4} = 2\). Thus, the lengths of the axes are 10 (horizontal) and 4 (vertical).
03

Find the Vertices

The vertices of the ellipse are determined from the center \((-2, 3)\), and since the major axis is horizontal, you add and subtract \(a\) from the x-coordinate of the center. Thus, the vertices are at \((-2+5, 3)\) and \((-2-5, 3)\), which are \((3, 3)\) and \((-7, 3)\).
04

Calculate the Distance to the Foci

For an ellipse, the distance to the foci \(c\) is given by \(c = \sqrt{a^2 - b^2}\). Substituting the values, \(c = \sqrt{25 - 4} = \sqrt{21}\).
05

Find the Foci

Using the center \((-2, 3)\), the foci, being along the horizontal axis, are found by adding and subtracting \(c\) from the x-coordinate of the center. Hence, the foci are located at \((-2 + \sqrt{21}, 3)\) and \((-2 - \sqrt{21}, 3)\).
06

Sketch the Ellipse

To sketch the ellipse, draw an ellipse centered at \((-2, 3)\) with horizontal stretch 5 and vertical stretch 2. Plot the vertices at \((3, 3)\) and \((-7, 3)\) and the foci at approximately \((2.58, 3)\) and \((-6.58, 3)\). The major axis should visually confirm the calculations, and label the center, vertices, and foci on the graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Vertices
The vertices of an ellipse are crucial as they define the endpoints of the major axis. In the given ellipse equation \(\frac{(x+2)^2}{25}+\frac{(y-3)^2}{4}=1\), we identify the center at \((-2, 3)\). Since the major axis is horizontal (because 25, which is under the \((x+2)^2\) term, is larger than 4), the vertices will lie along this axis.
  • The standard form of an ellipse with a horizontal major axis is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(a^2\) is the larger denominator. In this case, \(a^2 = 25\), giving \(a = 5\).
  • To find the vertices, we move \(a\) units left and right from the center's x-coordinate. This results in \((-2+5, 3) = (3, 3)\) and \((-2-5, 3) = (-7, 3)\).
Understanding the vertices gives us a direct insight into the placement and stretch of the ellipse along its major axis.
Ellipse Foci
Foci are two points located along the major axis of the ellipse that are used to define its shape and properties. The distance to each focus from the center can be found using the formula \(c = \sqrt{a^2 - b^2}\), where \(a\) and \(b\) are the semi-major and semi-minor axes.
  • With our current ellipse \(c = \sqrt{25 - 4} = \sqrt{21}\) or approximately \(4.58\).
  • Since the major axis is horizontal, we find the foci by adjusting the center's x-coordinate by \(\pm c\): add \(\sqrt{21}\) for the right focus, and subtract \(\sqrt{21}\) for the left focus.
  • This positions the foci at \((-2 + \sqrt{21}, 3)\) and \((-2 - \sqrt{21}, 3)\).
Including these accurate points on your sketch of the ellipse will help verify the correctness of other calculated points like the vertices and the axes.
Major and Minor Axes
The major and minor axes of an ellipse are lines of symmetry essential for its geometry. The longer axis is the major axis, and the shorter one is the minor axis.
  • For an ellipse in standard form, the major axis is determined by the larger of the two denominators. In this example, \(\frac{(x+2)^2}{25}+\frac{(y-3)^2}{4}=1\), 25 is larger, identifying the major axis as horizontal.
  • The length of the major axis is \(2a\), where \(a = \sqrt{25} = 5\), leading to a total length of 10 units.
  • The minor axis, being perpendicular, is \(2b\) with \(b = \sqrt{4} = 2\), giving a total length of 4 units.
The axes guide us to the broad and narrow stretches of the ellipse, providing visual symmetry and helping locate the vertices and foci accurately on a graph.
Ellipse Equation Standard Form
Understanding the standard form of an ellipse's equation is crucial in analyzing its geometric properties. The standard form is expressed as \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) represents the center of the ellipse.
  • In the given equation \(\frac{(x+2)^2}{25}+\frac{(y-3)^2}{4} = 1\), the center can be easily determined as \((-2, 3)\), derived from the transformations \(x+2 = 0\) and \(y-3 = 0\).
  • The larger denominator of 25 indicates that \(a^2 = 25\) corresponds to the horizontal stretch, establishing the major axis parallel to the x-axis.
  • The smaller denominator 4 means \(b^2 = 4\), denoting the ellipse stretches vertically only 2 units from the center.
Grasping the standard form enables us to find all crucial components such as vertices, foci, and lengths of axes, making it simpler to sketch and navigate the ellipse's properties.

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