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Chapter 12: Problem 39

Find an equation of the following ellipses and hyperbolas, assuming the center is at the origin. An ellipse whose major axis is on the \(x\) -axis with length 8 and whose minor axis has length 6

Short Answer

Expert verified
Answer: \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)

Step by step solution

01

Determine the semi-major axis and semi-minor axis

Since the length of the major axis is 8 and the minor axis is 6, we can find the semi-major axis and semi-minor axis by dividing the lengths by 2. So \(a = \frac{8}{2} = 4\), and \(b = \frac{6}{2} = 3\).
02

Write the standard equation of ellipse

Now that we have the values of \(a\) and \(b\), we can write the equation of the ellipse in standard form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2}=1\). Substitute the values of \(a\) and \(b\) into the equation and get \(\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1\).
03

Simplify the equation

Now we can simplify the equation by writing out the terms in the denominator: \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). This is the equation of the ellipse with the given conditions. The answer is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).

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Key Concepts

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

Semi-major axis
In any ellipse, the longest diameter is known as the major axis. It's quite simply the line segment passing through the center of the ellipse, reaching its widest points on either side. To find the semi-major axis, which is just half of this major axis, you simply divide the total length by 2.

In our exercise, the major axis is along the x-axis with a length of 8. So to find the semi-major axis, we perform the calculation: \[a = \frac{8}{2} = 4\]
  • This means you'll have a semi-major axis of 4 units.
  • It is aligned with the x-axis, as specified in the problem.
Understanding these basic properties helps define one half of the ellipse equation, particularly in determining how wide this stretched-out circle appears along the horizontal axis.
Semi-minor axis
The semi-minor axis serves a similar role to the semi-major axis, but for the shorter breadth of the ellipse. It is half of the minor axis and is usually perpendicular to the major axis within an ellipse.

For our particular problem, the minor axis length given is 6. To find the semi-minor axis, divide the minor axis length by 2:\[b = \frac{6}{2} = 3\]
  • This calculation shows the semi-minor axis is 3 units long.
  • For this ellipse, it lies along the y-axis, since the major axis is along the x-axis.
Gaining a clear grasp of how these axes influence the overall shape and proportions of the ellipse is crucial for performing further calculations and understanding the ellipse's geometric properties.
Standard form equation
Once you have identified the semi-major and semi-minor axes of an ellipse, the next step is to write its equation in standard form. This equation is fundamental in expressing the ellipse's geometric characteristics mathematically.

The standard form for the equation of an ellipse centered at the origin is given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2}=1\]where:
  • \(a\) is the length of the semi-major axis
  • \(b\) is the length of the semi-minor axis
For the exercise, the semi-major axis is 4, so \(a^2 = 16\). The semi-minor axis is 3, thus \(b^2 = 9\). Plugging these into the standard form equation, you get:\[\frac{x^2}{16} + \frac{y^2}{9} = 1\]This equation concisely describes the size and shape of the ellipse, showing how each point \((x, y)\) on the ellipse satisfies this relationship. Mastery of forming this standard equation is key to solving many problems related to ellipses.

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Most popular questions from this chapter

Let \(C\) be the curve \(x=f(t)\), \(y=g(t),\) for \(a \leq t \leq b,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous on \([a, b]\) and C does not intersect itself, except possibly at its endpoints. If \(g\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(x\)-axis is $$S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$. Likewise, if \(f\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(y\)-axis is $$S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$ (These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve \(y=f(x)\).) Use the parametric equations of a semicircle of radius \(1, x=\cos t\) \(y=\sin t,\) for \(0 \leq t \leq \pi,\) to verify that surface area of a unit sphere is \(4 \pi\).

Use a graphing utility to graph the hyperbolas \(r=\frac{e}{1+e \cos \theta}\) for \(e=1.1,1.3,1.5,1.7\) and 2 on the same set of axes. Explain how the shapes of the curves vary as \(e\) changes.

Slopes of tangent lines Find all points at which the following curves have the given slope. $$x=4 \cos t, y=4 \sin t ; \text { slope }=1 / 2$$

Assume \(m\) is a positive integer. a. Even number of leaves: What is the relationship between the total area enclosed by the \(4 m\) -leaf rose \(r=\cos (2 m \theta)\) and \(m ?\) b. Odd number of leaves: What is the relationship between the total area enclosed by the \((2 m+1)\) -leaf rose \(r=\cos ((2 m+1) \theta)\) and \(m ?\)

Find the length of the following polar curves. The three-leaf rose \(r=2 \cos 3 \theta\)
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