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Chapter 10: Problem 16

Ellipses Centered at (0, 0) Graph. $$9 x^{2}+16 y^{2}=144$$

Short Answer

Expert verified
The given ellipse has a semi-major axis of 4 units and a semi-minor axis of 3 units. Sketch it centered at (0,0) with vertices at (卤4,0) and co-vertices at (0,卤3).

Step by step solution

01

Identify the equation of the ellipse

The given equation of the ellipse is: \[ 9x^2 + 16y^2 = 144 \]This is a standard form of the ellipse equation.
02

Rewrite the equation in standard form

Divide both sides of the equation by 144 to rewrite it in the standard form of an ellipse equation: \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \]Simplify the fractions: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
03

Identify the values of a虏 and b虏

In the standard equation of an ellipse centered at (0,0): \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]Compare the given equation to this form to find: \[ a^2 = 16 \quad \text{and} \quad b^2 = 9 \]Thus, \( a = 4 \) and \( b = 3 \).
04

Determine the lengths of the axes

In an ellipse, the lengths of the semi-major axis (along the x-axis) and the semi-minor axis (along the y-axis) are given by values of \( a \) and \( b \) respectively. So, the lengths are: \( 2a = 8 \) and \( 2b = 6 \).
05

Sketch the ellipse

Draw the x and y axes, label the center at (0,0). Plot the vertices at (卤4, 0) and the co-vertices at (0, 卤3). Draw the shape of the ellipse using these points.

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Key Concepts

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

standard form of an ellipse
To understand an ellipse, it's important to begin with its equation. The standard form of the equation of an ellipse centered at the origin (0,0) is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]. Here, \(a\) and \(b\) are constants that determine the shape and size of the ellipse.

The term \(a^2\) is associated with the x-axis, and \(b^2\) is related to the y-axis. When this equation is satisfied, any pair of values for \(x\) and \(y\) will lie on the ellipse.
semi-major axis
In the context of an ellipse, the semi-major axis is a core concept. It represents half of the ellipse's longest diameter.

In the standard ellipse equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \], \(a\) corresponds to the semi-major axis. This means \(a\) is the distance from the center of the ellipse (0,0) to the farthest points along the x-axis. These points are called the 'vertices' of the ellipse.

Given our equation from the exercise, after simplification, \( a^2 = 16 \), giving us \( a = 4 \). Thus, the semi-major axis is 4 units long.
semi-minor axis
While the semi-major axis is the ellipse's longest radius, the semi-minor axis is its shortest. These two axes create the framework of the ellipse.

In the standard equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \], \(b\) represents the semi-minor axis. This measurement extends from the center of the ellipse to the closest edges along the y-axis, known as 'co-vertices' of the ellipse.

For our particular equation, we determined that \( b^2 = 9 \), which results in \( b = 3 \). Therefore, the semi-minor axis is 3 units long.
graphing ellipses
Graphing an ellipse might seem challenging at first, but it's straightforward once you understand the process.

Start by plotting the center of your ellipse at the origin (0,0).
Create marks at the vertices (卤4, 0) on the x-axis.
Then, mark the co-vertices on the y-axis at (0, 卤3).

These vertices and co-vertices act as visual guides.
Using them, sketch out the ellipse shape smoothly by connecting these points in a curved, oval shape. Ensure the ellipse stretches longer along the x-axis than the y-axis, reflecting the lengths of the semi-major and semi-minor axes respectively.

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Most popular questions from this chapter

Simplify. Assume that all variables represent positive numbers. $$\frac{\sqrt{200 x^{4} w^{2}}}{\sqrt{2 w}}$$

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. $$ 4 x^{2}-y^{2}+24 x+4 y+28=0 $$

Simplify. $$\sqrt{500}$$

Solve. $$x^{2}+4 x=60$$

Solve. $$\frac{4}{x+2}+\frac{3}{2 x-1}=2$$
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