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Chapter 11: Problem 25

Find the center, vertices, and foci of each ellipse and graph it. $$x^{2}+y^{2}=16$$

Short Answer

Expert verified
Center: (0,0). Vertices: (4,0), (-4,0), (0,4), (0,-4). Foci: (0,0)

Step by step solution

01

Identify the General Form

The given equation is \(x^2 + y^2 = 16\). This is already in the form of the standard ellipse equation \(Ax^2 + By^2 = C\). Recognize that it can be written as \(\frac{x^2}{16} + \frac{y^2}{16} = 1\), which fits the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(h = 0\), \(k = 0\), \(a^2 = 16\), and \(b^2 = 16\).
02

Determine the Center

The center of the ellipse is at the point \((h, k)\). Since \(h = 0\) and \(k = 0\), the center of the ellipse is at \((0, 0)\).
03

Calculate the Lengths of the Major and Minor Axes

Since \(a^2 = 16\) and \(b^2 = 16\), we find that \(a = \sqrt{16} = 4\) and \(b = \sqrt{16} = 4\). Because \(a = b\), the ellipse is actually a circle.
04

Find the Vertices

For a circle centered at \((0, 0)\) with radius 4, the vertices are located at the endpoints of the diameters along the axes. These points are \((4, 0)\), \((-4, 0)\), \((0, 4)\), and \((0, -4)\).
05

Determine the Foci

For a circle, the foci coincide with the center because \(c^2 = a^2 - b^2\) gives \(c^2 = 16 - 16 = 0\), resulting in \(c = 0\). Hence, the foci are at the center \((0, 0)\).
06

Graph the Circle

Plot the center at \((0, 0)\). Then, plot the vertices at \((4, 0)\), \((-4, 0)\), \((0, 4)\), and \((0, -4)\). Draw a circle that passes through these vertices.

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Key Concepts

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

center of ellipse
The center of an ellipse is a crucial point. It is the midpoint from which all distances to the ellipse are measured. For the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the center is represented by \(h\) and \(k\).
In our example \((x^2 + y^2 = 16)\), the center is at the origin \((0, 0)\) because \(h = 0\) and \(k = 0\).
The center helps to locate the symmetry of the ellipse.
major and minor axes
The major and minor axes are essential elements of an ellipse. The major axis is the longest diameter, and the minor axis is the shortest.
They intersect at the center of the ellipse. For our equation, \(\frac{x^2}{16} + \frac{y^2}{16} = 1\), \[a^2 = 16 \text{ and } b^2 = 16\], giving \[a = b = 4\].
Since \((a = b),\) our ellipse is actually a circle with equal axes. Each axis stretches 4 units away from the center, in all four directions.
vertices of ellipse
Vertices are the endpoints of the major and minor axes. Because a circle has equal axes, the vertices lie on both the horizontal and vertical axes at a distance equal to the radius.
For our circle \((x^2 + y^2 = 16)\), with radius 4, the vertices are:
  • \((4, 0)\)
  • \((-4, 0)\)
  • \((0, 4)\)
  • \((0, -4)\)
These points are symmetrically placed around the center.
foci of ellipse
The foci are two special points from which the sum of distances to any point on the ellipse remains constant. For ellipses, they help define the shape.
The distance to the foci is calculated by \[c^2 = a^2 - b^2\]. In our circle, since \[a = b, \text{ we get} \ c = 0\].
Hence, the foci coincide with the center: \((0, 0)\).
circle
A circle is a special type of ellipse where both axes are equal \((a = b)\). This makes every point on the circle equidistant from the center.
Our equation \((x^2 + y^2 = 16)\) confirms it's a circle with radius 4.
The center is at \((0, 0)\), the vertices are at the endpoints of the radius: \((4,0), (-4,0), (0,4)\) and \((0,-4)\).
Hence, a circle is a perfectly symmetric, round shape centered at \((0,0)\).

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Most popular questions from this chapter

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