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Conics Problem Part 3

  1. Jun 12, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    On picture above


    3. The attempt at a solution
    Again, I just want someone to check my work.
     
  2. jcsd
  3. Jun 12, 2008 #2
    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.
     
  4. Jun 13, 2008 #3
    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.
     
    Last edited: Jun 13, 2008
  5. Jun 13, 2008 #4
    For question 2, do you think I was right to mention the (h,k) values as a difference between the two graphs?
     
  6. Jun 13, 2008 #5
    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.
     
  7. Jun 13, 2008 #6
    We haven't used the terms focus or focii with these problems in class. What do you mean by them?
     
  8. Jun 13, 2008 #7
    Do you mean the center?
     
  9. Jun 13, 2008 #8
    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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