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HELP! Features of A Hyperbola CONIC!

  1. Apr 2, 2005 #1
    :cry: [tex] \frac {(x-1)^2} {9} - \frac {(y+2)^2} {25} = 1 [/tex]

    I think this is a vertical hyperbola
    with center (1,-2) a=5 b=3

    Transversal length=10
    Conjugate length =6

    Vertex (1,-7) and (1,3)
    Foci (1,sqrt(34)-2) and (1,-sqrt(34)-2)
    Asymptotes y+2=5/3(x-1) and y+2=-5/3 (x-1)

    I dont know if all these values are correct I need to find the intercepts if there are any can someone show me how?
  2. jcsd
  3. Apr 2, 2005 #2
    If you need to find the intercepts set x and y equal to zero and solve for the other point in each case. That's if you want to find the points (0,y) and (x,0)

    EDIT: Just to clarify, I meant you need set each variable equal to zero to find the intercepts separately, not at once.

    [tex] \frac {(0-1)^2} {9} - \frac {(y+2)^2} {25} = 1 [/tex]


    [tex] \frac {(x-1)^2} {9} - \frac {(0+2)^2} {25} = 1 [/tex]
    Last edited: Apr 2, 2005
  4. Apr 3, 2005 #3
    When I set y=0 I got x=-7 or x=7
    But when I set x=0 and solved for y I didnt get something as simple as x maybe I did something wrong, I think there are no y-intercepts am I correct?
    When I solve for y with x=0 I get an imaginary number
  5. Apr 3, 2005 #4
    Sketch the graph and think analytically about if there will be a x or y intercept. You have all of the information needed to draw the graph.
  6. Apr 3, 2005 #5
    Are all of my features correct? can someone please check!
  7. Apr 3, 2005 #6
    The ones I checked looked ok, I didn't look at them all though. Your center, transverse length, and conjugate length are correct. Show me where you think you messed up.
  8. Apr 3, 2005 #7
    for the yintercept I got

    y=+- Sqrt(-200) -2

    when I set x=0 and solved for y. Are my x-intercepts correct? Can you check my foci vertices and asymptotes too.
  9. Apr 3, 2005 #8


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    Nitpick: instead of saying "a=5, b=3", you should state to what those variables actually refer.

    This particular example is particularly important, because a lot of people (meaning me!) will immediately think of the usual form: [itex](x-h)^2/a^2 - (y-k)^2/b^2 = 1[/itex].
  10. Apr 3, 2005 #9
    I think ive made a huge mistake all of the features should be for the horizontal hyperbola not vertical hyperbola, right?
  11. Apr 3, 2005 #10
    indeed, it is a horizontal hyperbola (if by horizontal, you mean opens sideways).
  12. Apr 3, 2005 #11
    I have changed all my features but still having problems solving for the x-intercepts plz help me what do i do with the denominators?

    [tex] \frac {(x-1)^2} {9} - 4/25=1 [/tex]

    [tex] \frac {(x-1)^2} {9}=21/25 [/tex]

    I dont know what to do with the 9 what is the final answer?
  13. Apr 3, 2005 #12
    Multiply by 9 and you get:

    [tex](x-1)^2 = \frac{21}{25}*9[/tex]
    [tex]x - 1 = \sqrt {\frac{21}{25}*9}[/tex]
    [tex] x = \sqrt {\frac{21}{25}*9} + 1[/tex]

    Is this what you were having trouble with?
  14. Apr 3, 2005 #13
    mulpliply both sides by 9... bu what you actually mean is

    [tex]\frac{(x-1)^2}{9} - \frac{4}{25} = 1 \Longrightarrow \frac{(x-1)^2}{9} = \frac{29}{25}[/tex]

    so you can then just say

    [tex] (x-1)^2 = 9\frac{29}{25} \Longrightarrow x-1 = \pm\frac{3\sqrt{29}}{5} \Longrightarrow x = 1 \pm \frac{3\sqrt{29}}{5}[/tex]
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