Results 1 to 7 of 7

Math Help - Writing an ellipse in polar form a+ r cos(theta) = br.

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    Sydney
    Posts
    6

    Writing an ellipse in polar form a+ r cos(theta) = br.

    The given ellipse is (x-1)^2 /9 + y^2 /8 = 1

    Show that the equation of the given ellipse in polar coordinates has the form
    a + r cos(theta) = b r and find the values of a and b.

    This is an ellipse with x-intercepts at -2 and 4, and centre at (1,0).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    Have you at least attempted substituting \displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*} and \displaystyle \begin{align*} y = r\sin{(\theta)} \end{align*}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2013
    From
    Sydney
    Posts
    6

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    Yes, I got to 8r^2 cos^2(theta) -16rcos(theta) +9r^2 sin^2(theta) = 64, but i'm not sure how to get this into the correct form.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    Now notice that

    \displaystyle \begin{align*} 8r^2 \cos^2{(\theta)} + 9r^2 \sin^2{(\theta)} &= 8r^2 \cos^2{(\theta)} + 8r^2 \sin^2{(\theta)} + r^2\sin^2{(\theta)} \\ &= 8r^2 \left[ \cos^2{(\theta)} + \sin^2{(\theta)} \right] +r^2 \sin^2{(\theta)}  \end{align*}

    Does this look more manageable now?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2013
    From
    Sydney
    Posts
    6

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    Ok, so I reduced that to 8r^2 + r^2(sin^2 theta) = 64

    then taking square roots and rearranging,
    sqrt(8)r = -8 + rsin(theta)

    But this is in terms of sin and not cos, like the question says. Can this be solved by letting x= rsin(theta) and y = rcos(theta) [the opposite substitution as before] so that the final answer is in terms of cos(theta)?

    [EDIT] I forgot about the extra -16rcos(theta) term.
    So, using x=rcos(theta) and y = rsin(theta) this gives

    8r^2 + r^2 sin^2 (theta) -16rcos(theta) = 64. But how can I get rid of that extra rcos(theta) term?
    Last edited by rmca11; May 18th 2013 at 08:44 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    834
    Thanks
    209

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    8r^2 + r^2 sin^2 (theta) -16rcos(theta) = 64.
    8r^2 + r^2 ( 1-cos^2 (theta)) -16rcos(theta) = 64.
    8r^2 + r^2 - r^2 cos^2 (theta) -16rcos(theta) = 64.
    9r^2 - r^2 cos^2 (theta) -16rcos(theta) = 64.
    9r^2 - 64 = r^2 cos^2 (theta) +16rcos(theta)
    9r^2 - 64 = r^2 cos^2 (theta) +16rcos(theta) + 64 - 64
    9r^2 = [ rcos (theta) +8 ]^2
    Now you can take it further
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2013
    From
    Sydney
    Posts
    6

    Re: Writing an ellipse in polar form a+ r cos(theta) = br.

    Got it now, thank you both
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 28th 2013, 11:46 PM
  2. Replies: 3
    Last Post: April 10th 2013, 01:18 AM
  3. Replies: 1
    Last Post: November 16th 2010, 12:15 AM
  4. polar form of ellipse
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 3rd 2009, 08:15 AM
  5. Tilted ellipse in polar form
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 27th 2008, 07:35 AM

Search Tags


/mathhelpforum @mathhelpforum