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Chapter 10: Problem 13

Graph each inequality. See Examples 1 and \(2 .\) $$ \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1 $$

Short Answer

Expert verified
Graph an ellipse with center at origin and axes 2 and 3; shade inside including the boundary.

Step by step solution

01

Identify the Inequality's Form

The inequality \( \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1 \) is in the standard form of an ellipse. Specifically, it represents an ellipse centered at the origin \((0,0)\) with semi-major and semi-minor axes along the coordinate axes.
02

Determine the Axes Lengths

From the equation \( \frac{x^{2}}{4}+\frac{y^{2}}{9} = 1 \), we identify that the lengths of the semi-major and semi-minor axes are \( \sqrt{9} = 3 \) (along the y-axis) and \( \sqrt{4} = 2 \) (along the x-axis), respectively.
03

Sketch the Ellipse

Draw an ellipse centered at the origin \((0,0)\) going through the points \((2,0), (-2,0), (0,3), (0,-3)\), which are determined by the lengths of the semi-axes.
04

Graph the Inequality

Since the inequality is \( \leq \), shade the region inside the ellipse, including the boundary, because points inside satisfy the inequality. Use a solid line for the ellipse boundary to indicate the inclusion of points on the ellipse.

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Key Concepts

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

Graphing Inequalities
Graphing inequalities like \( \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1 \) involves more than just plotting a curve or shape. It requires understanding how to not only draw the boundary of a shape but also determine which area it encompasses based on the inequality sign. In our equation,
  • The symbol \( \leq \) indicates "less than or equal to."
  • This means that we're looking for all the points that lie inside or exactly on the ellipse.
Start by drawing the ellipse, which acts as the boundary. Then, because of the "equal to" part of "less than or equal to," shade the inside area. This shaded area shows every possible solution 鈥� every \( (x, y) \) pair that makes the inequality true. Use a solid line to draw the boundary to indicate that points on the ellipse are also included in the solution set.
Ellipse Equations
An ellipse equation, such as \( \frac{x^{2}}{4}+\frac{y^{2}}{9} = 1 \), is crucial in visualizing and understanding the shape's geometry. Ellipses are unique, oval-shaped figures defined clearly by their mathematical equation. Key features of an ellipse equation include:
  • The denominators, which determine the ellipse's width and height.
  • The numerator always being \( x^{2} \) and \( y^{2} \) at the origin.
The general form \( \frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1 \) outlines an ellipse centered at the origin. Here, the fraction forms represent the dilation effect on both axes, shaping the ellipse's form. For our inequality, the ellipse equation helps identify its size and orientation, playing a crucial role in correct plotting.
Standard Form of an Ellipse
The standard form of an ellipse provides a systematic way to write its equation. It looks like this: \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). The essence of this format is in its parts:
  • \( a^2 \) and \( b^2 \) are the squares of the lengths of the semi-axes.
  • The numerator \( x^2 \) and \( y^2 \) indicate the center is at the origin (\(0,0\)).
For our specific ellipse, \( a^2 = 4 \) and \( b^2 = 9 \), translating to semi-axis lengths of 2 and 3. The form immediately tells us the ellipse's dimensions and its center鈥檚 location. This standardized way of writing it enables easy understanding and allows us to quickly sketch or interpret the ellipse's features by plugging into this formula.
Axes Lengths
Axes lengths are fundamental in defining an ellipse鈥檚 size and shape. They are portrayed through the terms \( a \) and \( b \), representing half the lengths of the major and minor axes of the ellipse. To find them, you take the square roots of the denominators from the standard form equation:
  • For the x-axis, you'd determine \( a \) from \( a^2 \); here, \( a^2 = 4 \) so \( a = 2 \).
  • For the y-axis, \( b \) is from \( b^2 \); here, \( b^2 = 9 \) so \( b = 3 \).
These axes dictate how far the ellipse extends in each direction from its center. In this scenario, the semi-major axis is of length 3 along the y-axis, and the semi-minor axis is length 2 along the x-axis. This knowledge is essential when sketching as it ensures accuracy in portraying the ellipse's dimensions.

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Most popular questions from this chapter

Graph each inequality in two variables. $$ y<2 x-1 $$

Sketch the graph of each equation. \(\frac{(y+4)^{2}}{4}-\frac{x^{2}}{25}=1\)

The orbits of stars, planets, comets, asteroids, and satellites all have the shape of one of the conic sections. Astronomers use a measure called eccentricity to describe the shape and elongation of an orbital path. For the circle and ellipse, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=\left|a^{2}-b^{2}\right|\) and \(d\) is the larger value of a or b. For a hyperbola, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=a^{2}+b^{2}\) and the value of \(d\) is equal to a if the hyperbola has \(x\) -intercepts or equal to b if the hyperbola has \(y\) -intercepts. A. \(\frac{x^{2}}{36}-\frac{y^{2}}{13}=1\) B. \(\frac{x^{2}}{4}+\frac{y^{2}}{4}=1\) C. \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) D. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}=1\) G. \(\frac{x^{2}}{16}-\frac{y^{2}}{65}=1\) E. \(\frac{x^{2}}{17}+\frac{y^{2}}{81}=1\) F. \(\frac{x^{2}}{36}+\frac{y^{2}}{36}=1\) H. \(\frac{x^{2}}{144}+\frac{y^{2}}{140}=1\) What do you notice about the values of \(e\) for the equations you identified as hyperbolas?

The orbits of stars, planets, comets, asteroids, and satellites all have the shape of one of the conic sections. Astronomers use a measure called eccentricity to describe the shape and elongation of an orbital path. For the circle and ellipse, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=\left|a^{2}-b^{2}\right|\) and \(d\) is the larger value of a or b. For a hyperbola, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=a^{2}+b^{2}\) and the value of \(d\) is equal to a if the hyperbola has \(x\) -intercepts or equal to b if the hyperbola has \(y\) -intercepts. A. \(\frac{x^{2}}{36}-\frac{y^{2}}{13}=1\) B. \(\frac{x^{2}}{4}+\frac{y^{2}}{4}=1\) C. \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) D. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}=1\) G. \(\frac{x^{2}}{16}-\frac{y^{2}}{65}=1\) E. \(\frac{x^{2}}{17}+\frac{y^{2}}{81}=1\) F. \(\frac{x^{2}}{36}+\frac{y^{2}}{36}=1\) H. \(\frac{x^{2}}{144}+\frac{y^{2}}{140}=1\) For each of the equations \(\mathrm{A}-\mathrm{H}\), calculate the eccentricity \(e .\)

Find each function value if \(f(x)=3 x^{2}-2 .\) See Section 3.2 $$ f(b) $$
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