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Chapter 11: Problem 7

Graph the ellipses. In case, specify the lengths of the major and minor axes, the foci, and the eccentricity. For Exercises \(13-24,\) also specify the center of the ellipse. $$16 x^{2}+9 y^{2}=144$$

Short Answer

Expert verified
The ellipse center is (0, 0); major axis is 8, minor axis is 6, foci at (0, 卤鈭�7), and eccentricity is 鈭�7/4.

Step by step solution

01

Write the given equation in standard form

Start with the equation of the ellipse: \(16x^2 + 9y^2 = 144\). Divide every term by 144 to help rewrite the equation in a format that matches the standard form of an ellipse. This will yield: \(\frac{x^2}{9} + \frac{y^2}{16} = 1\).
02

Identify lengths of major and minor axes

In the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the larger denominator corresponds to \(b^2\) since \(16 > 9\). Thus, \(b=4\) (major axis), and \(a=3\) (minor axis). Length of the major axis: \(2b = 8\); Length of the minor axis: \(2a = 6\).
03

Find the foci of the ellipse

The foci are located at a distance \(c\) from the center along the major axis. Calculate \(c\) using the relationship \(c^2 = b^2 - a^2\). That is: \(c^2 = 16 - 9\), \(c = \sqrt{7}\). The foci are at \((0, \pm \sqrt{7})\).
04

Determine the center of the ellipse

Since the equation, \(\frac{x^2}{9} + \frac{y^2}{16} = 1\), is already centered at the origin \((0, 0)\), the center of the ellipse is \((0, 0)\).
05

Calculate the eccentricity of the ellipse

Eccentricity \(e\) is given by \(e = \frac{c}{b}\). Substitute the known values: \(e = \frac{\sqrt{7}}{4}\). This measures the ellipse's deviation from being a circle.

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Key Concepts

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

Ellipses Standard Form
To graph an ellipse effectively, it's crucial to first convert its equation into the **standard form**. The general standard form for an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. An important point to notice is the sum equals 1, which simplifies comparisons to the equation's constants.

For the given exercise, we start with the equation \(16x^2 + 9y^2 = 144\). To rewrite this in the standard form, divide each term by 144, resulting in \(\frac{x^2}{9} + \frac{y^2}{16} = 1\). This structure helps us to better identify the axis lengths and other properties of the ellipse. Understanding this transformation is key to unlocking the geometric features of the ellipse.
Major and Minor Axes
In an ellipse, the **major axis** is the longest diameter that passes through the center, while the **minor axis** is the shorter one. Determining which one is which in the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) depends on the denominators.

For the rewritten equation \(\frac{x^2}{9} + \frac{y^2}{16} = 1\), since 16 is greater than 9, 16 corresponds to \(b^2\), making \(b = 4\) the semi-major axis. Conversely, \(a = 3\) represents the semi-minor axis.

The full lengths of these axes are calculated as follows:
  • Length of the major axis: \(2b = 8\)
  • Length of the minor axis: \(2a = 6\)
Knowing these lengths is vital as they dictate the ellipse's width and height on a graph.
Foci of Ellipses
The **foci** of an ellipse are two special points equidistant from the center along the major axis. They help in figuring out how "stretched" the ellipse is. In mathematical terms, the distance \(c\) from the center to each focus is found by \(c^2 = b^2 - a^2\).

For our example:
  • \(c^2 = 16 - 9 = 7\)
  • \(c = \sqrt{7}\)
The foci are located vertically at \((0, \pm \sqrt{7})\), because the major axis is along the y-axis. Understanding the location of the foci gives you insight into the ellipse's direction and orientation.
Eccentricity
**Eccentricity** is a measure showing how much an ellipse deviates from being a perfect circle. It ranges from 0 to 1, where 0 represents a circle. For an ellipse, it's calculated using \( e = \frac{c}{b} \), with \(c\) being the distance to the foci and \(b\) representing the semi-major axis.

Applying this formula to our ellipse:
  • \(e = \frac{\sqrt{7}}{4}\)
This eccentricity indicates the shape's elongation. The closer the eccentricity is to 0, the more circular it is, while values closer to 1 indicate it is more stretched. Grasping eccentricity helps in predicting an ellipse's shape and how it might appear visually when graphed.

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Most popular questions from this chapter

In this exercise we consider how the eccentricity \(e\) influences the graph of an ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1.\) (a) For simplicity, we suppose that \(a=1\) so that the equation of the ellipse is \(x^{2}+\left(y^{2} / b^{2}\right)=1 .\) Solve this equation for \(y\) to obtain $$y=\pm b \sqrt{1-x^{2}}$$ (b) Assuming that \(a=1,\) show that \(b\) and \(e\) are related by the equation \(b^{2}=1-e^{2},\) from which it follows that \(b=\pm \sqrt{1-e^{2}} .\) The positive root is appropriate here because \(b>0 .\) Thus, we have $$b=\sqrt{1-e^{2}}$$ (c) Using equation (2) to substitute for \(b\) in equation (1) yields \(y=\sqrt{1-e^{2}} \sqrt{1-x^{2}} \quad\) or \(\quad y=-\sqrt{1-e^{2}} \sqrt{1-x^{2}}\) This pair of equations represents an ellipse with semimajor axis 1 and eccentricity \(e .\) Using the value \(e=0.3,\) graph equations ( 3 ) in the viewing rectangle [-1,1] by [-1,1] Use true proportions and, for comparison, add to your picture the circle with radius 1 and center \((0,0) .\) Note that the ellipse is nearly circular. (d) Follow part (c) using \(e=0.017 .\) This is approximately the eccentricity for Earth's orbit around the Sun. How does the ellipse compare to the circle in this case? (e) Follow part (c) using, in turn, \(e=0.4, e=0.6, e=0.8\) \(e=0.9, e=0.99,\) and \(e=0.999 .\) Then, in complete sentences, summarize what you've observed.

Graph the equations. $$x^{2}+4 x y+4 y^{2}=1$$

This exercise outlines the steps required to show that the equation of the tangent to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) on the ellipse is \(\left(x_{1} x / a^{2}\right)+\left(y_{1} y / b^{2}\right)=1\) (a) Show that the equation \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is equivalent to $$b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$$(1) Conclude that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse if and only if $$b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}$$(2) (b) Subtract equation ( \(2)\) from equation (1) to show that $$b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0$$(3) Equation ( 3 ) is equivalent to equation ( 1 ) provided only that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse. In the following steps, we will find the algebra much simpler if we use equation ( 3 ) to represent the ellipse, rather than the equivalent and perhaps more familiar equation (1). (c) Let the equation of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) be $$y-y_{1}=m\left(x-x_{1}\right)$$ Explain why the following system of equations must have exactly one solution, namely, \(\left(x_{1}, y_{1}\right);\) $$\left\\{\begin{array}{ll}b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0 & (4) \\ y-y_{1}=m\left(x-x_{1}\right) & (5)\end{array}\right.$$ (d) Solve equation ( 5 ) for \(y\), and then substitute for \(y\) in equation (4) to obtain \(b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2} m^{2}\left(x-x_{1}\right)^{2}+2 a^{2} m y_{1}\left(x-x_{1}\right)=0\) (e) Show that equation (6) can be written \(\left(x-x_{1}\right)\left[b^{2}\left(x+x_{1}\right)+a^{2} m^{2}\left(x-x_{1}\right)+2 a^{2} m y_{1}\right]=0\) (f) Equation (7) is a quadratic equation in \(x,\) but as was pointed out earlier, \(x=x_{1}\) must be the only solution. (That is, \(x=x_{1}\) is a double root.) Thus the factor in brackets must equal zero when \(x\) is replaced by \(x_{1}\) Use this observation to show that $$m=-\frac{b^{2} x_{1}}{a^{2} y_{1}}$$ This represents the slope of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) (g) Using this value for \(m,\) show that equation (5) becomes $$b^{2} x_{1} x+a^{2} y_{1} y=b^{2} x_{1}^{2}+a^{2} y_{1}^{2}$$(8) (h) Now use equation (2) to show that equation (8) can be written $$\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$$ which is what we set out to show.

If \(\overline{P Q}\) is a focal chord of the parabola \(x^{2}=4 p y\) and the coordinates of \(P\) are \(\left(x_{0}, y_{0}\right),\) show that the coordinates of \(Q\) are $$\left(\frac{-4 p^{2}}{x_{0}}, \frac{p^{2}}{y_{0}}\right)$$

If \(\overline{P Q}\) is a focal chord of the parabola \(y^{2}=4 p x\) and the coordinates of \(P\) are \(\left(x_{0}, y_{0}\right),\) show that the coordinates of \(Q\) are $$\left(\frac{p^{2}}{x_{0}}, \frac{-4 p^{2}}{y_{0}}\right)$$
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