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Chapter 10: Problem 23

Ellipses Centered at (0, 0) Graph. $$16 x^{2}=16-y^{2}$$

Short Answer

Expert verified
The equation \(x^2 + \frac{y^2}{16} = 1\) represents an ellipse centered at (0, 0) with a semi-major axis of 4 and a semi-minor axis of 1.

Step by step solution

01

Rewrite the Equation

Start by rewriting the given equation in the standard form of an ellipse. The given equation is \(16x^{2}=16-y^{2}\). Move all terms to one side to set the equation to zero: \(16x^{2} + y^{2} = 16\).
02

Divide by 16

To get the equation into the standard form of an ellipse, divide every term by 16: \(\frac{16x^2}{16} + \frac{y^2}{16} = \frac{16}{16}\). This simplifies to \(x^2 + \frac{y^2}{16} = 1\).
03

Identify the Components of the Ellipse

Recognize that in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), \(a^2 = 1\) and \(b^2 = 16\), yielding \(a = 1\) and \(b = 4\). This tells us that the ellipse is centered at (0, 0), has a semi-major axis length of 4 along the y-axis, and a semi-minor axis length of 1 along the x-axis.
04

Sketch the Ellipse

Plot the points where the ellipse intersects the axes. These points are (1, 0), (-1, 0), (0, 4), and (0, -4). Draw the ellipse by smoothly connecting these points while maintaining an oval shape.

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Key Concepts

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

Standard Form of Ellipse
To understand ellipses in algebra, it's crucial to know their standard form. The equation of an ellipse centered at the origin (0,0) looks like this:
\[{\frac{x^2}{a^2}} + {\frac{y^2}{b^2}} = 1\].
Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. By transforming the given equation, \(16x^2 + y^2 = 16\), into this standard form, we can better understand the shape and orientation of the ellipse.
Graphing Ellipses
Graphing an ellipse involves plotting its major and minor axes and connecting these points smoothly. For the given equation, \(x^2 + {\frac{y^2}{16}} = 1\), we identify four crucial points: (1, 0), (-1, 0), (0, 4), and (0, -4). These points are where the ellipse intersects the x and y axes.
  • Identify where the ellipse intersects the axes.
  • Plot these points on the graph.
  • Draw the oval shape smoothly connecting these points.
This makes visualizing the ellipse easier and ensures accuracy.
Semi-Major and Semi-Minor Axes
The semi-major and semi-minor axes are fundamental when graphing an ellipse. They are the distances from the center to the farthest and shortest points of the ellipse, respectively. For the given equation, \(x^2 + {\frac{y^2}{16}} = 1\), we find:
  • Semi-major axis (longest): \(b = 4\), along the y-axis.
  • Semi-minor axis (shortest): \(a = 1\), along the x-axis.
Understanding these helps plot the ellipse accurately and emphasizes the importance of these axes in defining the ellipse鈥檚 shape.
Centering Ellipse at Origin
Ellipses are often centered at the origin (0, 0) for simplicity. When centered at the origin, the equations become easier to handle, as seen in our case with \(x^2 + {\frac{y^2}{16}} = 1\). This property allows a symmetric plot around both the x and y-axes. Consider the following steps:
  • Identify the standard form of the ellipse centered at the origin.
  • Recognize symmetry about the center.
  • Plot axes intersections and construct the ellipse.
Centering at the origin simplifies understanding and graphing ellipses.

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Most popular questions from this chapter

Solve. Art. Elliot is designing a rectangular stained glass miniature that has a perimeter of 28 cm and a diagonal of length 10 cm. What should the dimensions of the glass be?

If an equation has two variable terms with the same degree, can its graph be a parabola? Why or why not?

Find an equation of an ellipse that contains the following points. $$(-2,-1),(6,-1),(2,-4), \text { and }(2,2)$$

Find an equation of an ellipse that contains the following points. $$(-7,0),(7,0),(0,-10), \text { and }(0,10)$$

The standard form for equations of horizontal or vertical hyperbolas centered at \((h, k)\) are as follows: $$ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 $$ (Graph can't copy) $$ \frac{(y-k)^{2}}{b^{2}}-\frac{(x-h)^{2}}{a^{2}}=1 $$ The vertices are as labeled and the asymptotes are $$ y-k=\frac{b}{a}(x-h) \text { and } y-k=-\frac{b}{a}(x-h) $$ For each of the following equations of hyperbolas, complete the square, if necessary, and write in standard form. Find the center, the vertices, and the asymptotes. Then graph the hyperbola. $$ 25(x-4)^{2}-4(y+5)^{2}=100 $$
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