MATH 155 SUPPLEMENTAL NOTES 30
CLASSIFYING CONIC SECTIONS BY ECCENTRICITY
ECCENTRICITY
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Let us investigate the behavior of c and a for the following ellipses. |
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Equation |
a |
c |
Color of graph |
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2 |
0 |
Red |
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2 |
1 |
Blue |
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2 |
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Green |
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2 |
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Purple |
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2 |
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Pink |
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2 |
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Grey |
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2 |
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Light Green |
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2 |
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Black |
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Now look at the graph of all of these ellipses on the same axis. When c = 0, we have a circle (red), and as c approaches the value of a, the ellipses become more elongated. So as c increases, the ellipse will flatten out until c = a. When c = a, the ellipse will degenerate to a line segment. The ratio of c to a is used to describe the various shapes an ellipse can assume. This ratio is called the eccentricity of the ellipse. It has one major use outside of calculus, and it is the fact that it is used to describe the orbits of the planets around the sun. As an example, Venus's orbit is almost circular, whereas Pluto's is highly elongated. |
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FACT: |
The eccentricity of the ellipse
is
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FACT: |
The eccentricity of the hyperbola
is
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In both the ellipse and the hyperbola, the eccentricity is defined to be the ratio of c to a, or the distance between the foci to the distance between the vertices. |
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FACT: |
If e = 0, then the conic section is a circle. If e < 1, then the conic section is an ellipse. If e = 1, then the conic section is a parabola. If e > 1, then the conic section is a hyperbola. |
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Now, let us determine the eccentricities, foci, and other relevant |
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EXAMPLE 1: |
Find the eccentricity, foci, and directrices for the ellipse 9x 2 + 10y 2 = 90. |
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SOLUTION: |
First I will put the equation into standard form by dividing everything by 90.
From this equation we can determine the value of a.
Now to determine the value of c.
The eccentricity for this ellipse is the following.
The foci for this ellipse are (± 1, 0). The directrices of this ellipse are of the form x = ± a / e. Therefore, the directrices for this ellipse are the following equations.
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EXAMPLE 2: |
Find the eccentricity, foci, and directrices for the hyperbola y 2 - 3x 2 = 3. |
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SOLUTION: |
First, I will put this equation into standard form by dividing all sides by 3.
Now I will find the center-to-focus distance c.
From the above equation I can determine the value for a.
The eccentricity for this hyperbola is the following.
The foci are on the y-axis, so the foci are (0, ± 2). The directrices for this hyperbola are of the form y = ± a / e, so the directrices for this hyperbola are the following equations.
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EXAMPLE 3: |
Given that the foci are at (0, ± 3) and the eccentricity is 0.5, find the equation of the ellipse centered at the origin. |
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SOLUTION: |
Given that the ellipse is centered at the origin and that the foci are (0, ± 3), this implies that the foci are on the y-axis and that c = 3. Recall that e = c/a, so using this fact we will determine the value for a.
Now we have to determine the value of b 2.
Using all the data that we have now determined, we can write the equation of the ellipse. This is the equation for this ellipse.
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EXAMPLE 4: |
Given the vertices (± 10, 0) and the eccentricity 0.24, find the equation of the ellipse centered at the origin. |
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SOLUTION: |
Given that the vertices are at (± 10, 0), this implies that they lie on the x-axis and a = 10. Given that the eccentricity, e =0.24, we can determine the value for c.
Now to determine b 2.
Here is the equation for this ellipse.
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EXAMPLE 5: |
Given the focus (4, 0) and the corresponding directrices x = 16/ 3, find the equation of the ellipse centered at the origin. |
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SOLUTION: |
Given one of the foci, (4, 0), this implies that the foci lie on the x-axis and that c = 4. Remember that the equation for the directrices is x = a / e, and that the equation for e = c/ a. Therefore ae = c or a = c/e. Using this fact, we can modify the equation for x.
Now to determine the value for e.
Now to find the value for a.
Now to find the value for b 2.
Here is the equation of this ellipse.
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EXAMPLE 6: |
Given the eccentricity, e = 3, and foci at (± 3, 0), find the equation of the hyperbola centered at the origin. |
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SOLUTION: |
With the foci being (± 3, 0), this implies that the foci lie on the x-axis. Recall, that e = c/a. I will use this equation to determine the value for a.
Now I will determine the value for b 2.
Finally, here is the equation for this hyperbola.
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EXAMPLE 7: |
Given the focus (4, 0) and its corresponding directrix x = 2, find the equation of the hyperbola centered at the origin. |
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SOLUTION: |
Since the foci is (± 4, 0), this tells me that the foci lie on the x-axis and c = 4. Now I have to find the value for a 2.
Now to determine the value of b 2.
Therefore, the equation of this hyperbola is the following.
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As you can see in this set of notes, we can use the eccentricity to determine what type of conic section we are dealing with, to help us construct the equation of the conic section. The concept of the eccentricity plays a part in astronomy. It describes the orbit of the planet, and it allows astronomers to calculate when comets will appear in our skies. Work through these examples, taking note of the different equations. As with any set of supplemental notes, if you have any questions on any of the examples, please feel free to contact me.
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