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Chapter 13: Problem 5

graph each ellipse. $$\frac{x^{2}}{25}+\frac{y^{2}}{64}=1$$

Short Answer

Expert verified
The graph of the given equation is an ellipse with a center at the origin (0,0) and vertices at (-5,0), (5,0), (0,-8) and (0,8).

Step by step solution

01

Identify the semi-major axis and semi-minor axis

Firstly, we see that the denominator under the x term is 25, which is \(a^{2}\). Therefore, a is 5. This means the ellipse extends 5 units to the left and 5 units to the right of the center. Similarly, the denominator under the y is 64, which is \(b^{2}\). So, b is 8. This means the ellipse extends 8 units up and 8 units down from the center.
02

Plot the center and the vertices of the ellipse

The center of the ellipse is at the origin (0,0) because there are no translation terms in the equation. The vertices of the ellipse are then at (卤a, 0) and (0, 卤b). Substituting the given values, the vertices are at (卤5, 0) and (0, 卤8). So, the four points to plot are (-5,0), (5,0), (0,-8), and (0,8).
03

Draw the ellipse

Now connect the vertices with a smooth, rounded curve to form the ellipse. The curve should bulge farther out in the y direction since the semi-major axis is the y-axis.

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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 a fundamental part of an ellipse. It is the longest radius and determines how far the ellipse stretches from its center. The semi-major axis is along the y-axis in this example 鈥� derived from the equation \( \frac{x^{2}}{25}+\frac{y^{2}}{64}=1 \). Here, \( b^{2} = 64 \) gives us \( b = 8 \). This tells us that the semi-major axis extends 8 units above and 8 units below the center of the ellipse.
  • The semi-major axis is always associated with the larger of the two denominators in the standard form of the ellipse equation.
  • It runs from the center out to the ellipse's top and bottom.
  • In this specific ellipse equation, the semi-major axis defines how vertically elongated the shape is.
    • Once identified, it plays a crucial role in sketching the ellipse accurately. Learners should always determine and verify the semi-major axis to ensure correct orientation and dimensions of the ellipse.
Semi-Minor Axis
On the other hand, the semi-minor axis is the shortest radius of an ellipse. It dictates the closer horizontal extent of the ellipse relative to its center. For the given equation, the semi-minor axis lies on the x-axis. This is shown by \( a^{2} = 25 \), where \( a = 5 \). This indicates that it stretches from the center to 5 units on both sides left and right.
  • The semi-minor axis is tied to the smaller denominator in the ellipse equation.
  • It helps determine how wide the ellipse appears horizontally.
  • Despite being labeled 'minor,' it is equally important for plotting the ellipse.
The semi-minor axis complements the semi-major axis, together defining the complete address of the ellipse in terms of both vertical and horizontal extents.
Ellipse Vertices
The vertices of an ellipse are key points that help in outlining its shape. These include the endpoints of both the semi-major and semi-minor axes. In our illustration, the vertices mark the interception of the ellipse with the axes. From the equation \( \frac{x^{2}}{25}+\frac{y^{2}}{64}=1 \):
  • For the semi-major axis (vertical), the vertices are at \( (0, 8) \) and \( (0, -8) \).
  • For the semi-minor axis (horizontal), the vertices are at \( (5, 0) \) and \( (-5, 0) \).
These four vertices are plotted to outline the initial quadrants of the ellipse. They aid in sketching the curve, ensuring that the ellipse is drawn symmetrically around its center. With vertices plotted, a smooth, continuous curve should connect these points to form the perfect ellipse, marking the exterior bounds as defined by the axes.

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Most popular questions from this chapter

What is a parabola?

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$x^{2}+4 y^{2}=16$$

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$x=-3(y-1)^{2}-2$$

Use the vertex and intercepts to sketch the graph of each equation. If needed, find additional points on the parabola by choosing values of y on each side of the axis of symmetry. $$x=-3(y+1)^{2}-2$$

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$3 x^{2}=27+3 y^{2}$$
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