Results 1 to 4 of 4

Thread: Ellipse

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Exclamation Ellipse

    Hi

    How exactly would I do this?

    The normal to the ellipse x^2/a^2 + y^2/b^2 = 1 at P(x1, y1) meets the x-axis in N and the Y-axis in G. Prove PN/NG = (1-e^2)/e^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    See this thread for some general comments about ellipses that should help with this problem. Remember that the eccentricity is given by e^2 = 1 - \frac{b^2}{a^2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Exclamation

    Hi

    I've spent a few days on this but I cant solve it. PLEASE HELP!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Sunyata View Post
    I've spent a few days on this but I cant solve it.
    I'm not entirely surprised. This is a longer and messier calculation than I would have guessed. Here is an outline of how to do it.

    Take P to be the point (a\cos\theta,b\sin\theta). Then (see the link in my other comment above) the normal at P has equation yb\cos\theta - xa\sin\theta = (b^2 - a^2)\cos\theta\sin\theta.

    Put y = 0 in that equation to see that N is the point \Bigl(\frac{(a^2-b^2)\cos\theta}a,0\Bigr). Put x = 0 to see that G is the point \Bigl(0,\frac{(b^2-a^2)\sin\theta}b\Bigr).

    Then use the usual distance formula to see that PN^2 = \Bigl(\frac{b^2\cos\theta}a\Bigr)^2 + b^2\sin^2\theta = \frac{b^2(b^2\cos^2\theta + a^2\sin^2\theta)}{a^2}.

    Similarly NG^2 = (a^2-b^2)^2\Bigl(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\Bigr) = \frac{(a^2-b^2)^2(b^2\cos^2\theta + a^2\sin^2\theta)}{a^2b^2}.

    Therefore \Bigl(\frac{PN}{NG}\Bigr)^2 = \frac{b^4}{(a^2-b^2)^2}, and so \frac{PN}{NG} = \frac{b^2}{a^2-b^2} = \frac{a^2(1-e^2)}{a^2e^2} = \frac{1-e^2}{e^2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ellipse
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 17th 2010, 06:48 AM
  2. ellipse
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 16th 2010, 11:29 AM
  3. ellipse 2
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 15th 2010, 11:41 AM
  4. ellipse
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jun 7th 2010, 04:08 PM
  5. Ellipse
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Mar 17th 2008, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum