Results 1 to 5 of 5

Math Help - Finding b-value of ellipse equation for given eccentricity

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    5

    Finding b-value of ellipse equation for given eccentricity

    Hi, this is my first time posting. I need help with this problem:

    Suppose the equation for an ellipse is given by

    \frac{(x-3)^2}{4^2}+\frac{(y+2)^2}{b^2}=1

    Find b so that the eccentricity is 0.75

    -------------------------------------------

    I know that eccentricity is defined as \frac{c}{a}, making c=3 and a=4. However, if I plug these values into the Pythagorean theorem, I get b^2=-7 (which is obviously incorrect, right? We're dealing with real numbers).

    How do I solve this problem? Thanks in advance.
    Last edited by bumpjump; May 6th 2010 at 03:55 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    You ran into trouble when you went from \frac{c}{a}=0.75 to c=3 and a=4. You don't know the actual values of c and a, just the ratio (you could have chosen c=6 and a=8 instead, and \frac{c}{a} would still be 0.75.

    Try substituting c=\sqrt{a^2-b^2}. Then you can go two ways (and get two solutions). Either substitute b=4 or a=4. If you think about it, there should be two solutions, a "tall, skinny" ellipse and a "short, fat" ellipse.

    So the first one will be \sqrt{a^2-4^2}=0.75a, and the second will be \sqrt{4^2-b^2}=0.75*4.

    I think you can find the solutions from there. If you have any further questions, please post again in this thread.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    5
    I'm not sure what you mean when you say substitute  c=\sqrt{a^2-b^2}. Should I be substituting a=4 into the equation to get c^2=4^2-b^2? What do I do from there? I missed this day of class and am trying to figure out how to do this but my textbook gives no examples.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by bumpjump View Post
    Hi, this is my first time posting. I need help with this problem:

    Suppose the equation for an ellipse is given by

    \frac{(x-3)^2}{4^2}+\frac{(y+2)^2}{b^2}=1

    Find b so that the eccentricity is 0.75

    -------------------------------------------

    I know that eccentricity is defined as \frac{c}{a}, making c=3 and a=4. However, if I plug these values into the Pythagorean theorem, I get b^2=-7 (which is obviously incorrect, right? We're dealing with real numbers).

    How do I solve this problem? Thanks in advance.
    I will point you to here: The Ellipse ........................................

    They have an example almost identical to your question!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    5
    Wow... I was wayyyy overthinking this problem. Thanks to both of you guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Eccentricity of ellipse
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 5th 2011, 04:33 AM
  2. Is the eccentricity of this ellipse correct?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 22nd 2009, 08:19 AM
  3. Replies: 3
    Last Post: August 9th 2009, 02:40 PM
  4. Replies: 2
    Last Post: December 19th 2008, 04:01 AM
  5. Finding equation of ellipse
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 11th 2007, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum