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Conics Problem Part 3

  • Thread starter temaire
  • Start date
  • #1
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Homework Statement


http://img103.imageshack.us/img103/3784/arch3tf0.jpg [Broken]​
[/URL]


Homework Equations


On picture above


The Attempt at a Solution


Again, I just want someone to check my work.
 
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Answers and Replies

  • #2
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For the first problem on this page, (again from the previous problem) for a hyperbola, [tex]c^2=a^2+b^2 [/tex]

c=distance from center to focii, a = distance from center to vertex
Incidentally, you don't need to calculate for a since it's given.
 
  • #3
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Whoops, I realized that the problem is using a different form of the vertical hyperbola instead of [tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex], so in this case switch a and b. In the other problem, where b=10 is correct.
 
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  • #4
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For question 2, do you think I was right to mention the (h,k) values as a difference between the two graphs?
 
  • #5
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Since (h,k) in the parabolic model is the vertex whereas the (h,k) in the semi-elliptical model is the center, I believe you should also compare the focus of the parabola and the focii of the ellipse.
 
  • #6
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We haven't used the terms focus or focii with these problems in class. What do you mean by them?
 
  • #7
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Do you mean the center?
 
  • #8
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The focii for an ellipse is the point that lies on the major axis(the longer side/axis) of the ellipse. There are two focii in this ellipse. (h+c,k) and (h-c,k). In an ellipse(for both vertial and horizontal ellipses), [tex]b^2=a^2-c^2[/tex], where a is always the large axis and b is the smaller axis.

In a parabola, the there is only one focus. Since this parabola opens down, then the focus is at (h,-c+k).
 

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