1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Least possible length

  1. Oct 21, 2013 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    If abscissa and ordinate of vertex of parabola [itex]y^2+2ax+2by-1=0[/itex] are equal then least possible length of latus rectum is

    2. Relevant equations

    3. The attempt at a solution
    The give equation can be rewritten as

    [itex]\left( y+b \right) ^2 = -2a\left\{ x-\dfrac{1+b^2}{2a} \right\}[/itex]

    As given in question

    [itex]-b=\dfrac{1+b^2}{2a} \\
    1+b^2+2ab=0[/itex]

    Length of latus rectum = a/2
    But what can be the range of values of a?
     
  2. jcsd
  3. Oct 21, 2013 #2
    You have "a" as a function of b, how about using that?
     
  4. Oct 21, 2013 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi utkarshakash! :smile:
    well, a can't be 0, can it? :wink:

    have you tried completing the square?​
     
  5. Oct 22, 2013 #4

    utkarshakash

    User Avatar
    Gold Member

    (1+b)^2+2b(a-1)=0

    But how does this help?
     
  6. Oct 22, 2013 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    a can obviously be 1

    is it possible eg for a to be 1/2 ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted