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Chapter 7: Problem 63

Describe how to locate the foci for \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

Short Answer

Expert verified
The foci of the ellipse defined by the equation \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) are located at the points (3, 0) and (-3, 0).

Step by step solution

01

Identify Semi-Major and Semi-Minor Axes

Compare the given equation with the general equation of an ellipse. The larger number in the denominator denotes the semi-major axis and the smaller one denotes the semi-minor. Here, \(a^{2}=25\) and \(b^{2}=16\). Thus, \(a=5\) and \(b=4\).
02

Calculate the Distance to the Foci

Now use the formula \(c=\sqrt{a^{2}-b^{2}}\) to calculate the distance of the foci from the center. Substituting \(a=5\) and \(b=4\) in the formula, we get \(c=\sqrt{5^{2}-4^{2}}=\sqrt{9}=3\).
03

Locate the Foci

Finally, count the calculated number of units away from the center along the larger of the axes to find the foci. In this case, as the larger axis is the x-axis, foci will be at (3, 0) and (-3, 0).

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Key Concepts

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

Understanding the Semi-Major Axis
An ellipse might seem complex, but it simplifies when you break it down into its parts, like the semi-major axis. This is the longest radius of an ellipse, reaching from the center to the edge across its widest part. For any equation of an ellipse, the greater denominator under the squared terms determines the semi-major axis.
  • For example, in the equation given \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), the denominator 25 identifies the semi-major axis.
  • The length of this axis is calculated using the square root of the denominator, 25, thus \(a=\sqrt{25}=5\).
Understanding this helps determine both the shape and orientation of the ellipse. If the larger denominator is under \(x^{2}\), the ellipse is oriented horizontally, as in our example.
Exploring the Semi-Minor Axis
The semi-minor axis is the shorter sibling to the semi-major axis. Its role is equally important but often less noticeable. It determines the width of the ellipse across its narrower part, stretching from the center outward to the edge. Let's see how it fits into the equation:
  • From the equation \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), the smaller denominator, 16, indicates the semi-minor axis.
  • To find the length, take the square root of 16, giving \(b=\sqrt{16}=4\).
Together with the semi-major axis, the semi-minor axis forms the elliptical shape by controlling the vertical stretch. So in a horizontally oriented ellipse like ours, \(b\) is crucial for determining the height from top to bottom.
Locating the Foci of an Ellipse
The foci of an ellipse are two special points located along its longest part, or the semi-major axis. They serve a unique purpose: the sum of distances from any point on the ellipse to the two foci is constant. Here鈥檚 how you locate them:
  • Begin by calculating the focal distance using the formula \(c=\sqrt{a^{2}-b^{2}}\). This gives you the distance from the center to each focus.
  • In our example, substituting \(a=5\) and \(b=4\) into the formula, you find \(c=\sqrt{5^2-4^2}=\sqrt{9}=3\).
  • Since the ellipse is oriented horizontally, place the foci by counting 3 units along the x-axis: one at (3,0) and the other at (-3,0).
Understanding the foci not only aids in defining the ellipse鈥檚 precise shape but also has applications in physics and engineering, where ellipses frequently appear.

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Most popular questions from this chapter

Graph each ellipse and give the location of its foci. $$\frac{(x+1)^{2}}{2}+\frac{(y-3)^{2}}{5}=1$$

Graph each ellipse and give the location of its foci. $$\frac{x^{2}}{25}+\frac{(y-2)^{2}}{36}=1$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$$

Graph each ellipse and give the location of its foci. $$\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1$$

Find the standard form of the equation of each ellipse satisfying the given conditions. $$\text { Foci: }(0,-4),(0,4) ; \text { vertices: }(0,-7),(0,7)$$
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