Results 1 to 5 of 5

Math Help - Ellipse Problems

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Exclamation Ellipse Problems

    Hi Could someone please give me hints for these last few questions I have to do....

    I'm really struggling to finish them off before the weekend's over

    Please, any help?
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Feb 2010
    Posts
    4
    Quote Originally Posted by xwrathbringerx View Post
    Hi Could someone please give me hints for these last few questions I have to do....

    I'm really struggling to finish them off before the weekend's over

    Please, any help?
    for the last one,
    i hope you know for an ellipse c^2+b^2=a^2.
    where 2c=focal distance, 2b=minor axis length, 2a=major axis length.
    So having origin as centre for the ellipse, the point (c,y) lies on the ellipse for some y which we will find.
    therefore, placing (c,y) in standard ellipse eqn:
    (c^2/a^2)+(y^2/b^2)=1.
    using c^2+b^2=a^2, we get y=b^2/a.
    thus distance of point from (c,0) i.e. a focus, is b^2/a which is the semi-latus rectum.

    in case u dont know c^2+b^2=a^2.
    from defn. of ellipse sum of distances of a point from two focii of an ellipse is equal to length of major axis. So, for a standard ellipse having origin as centre, for the point (0,b),
    dist of (0,b) from (-c,0)=dist of (0,b) from (c,0)=a. [from geometry]
    the triangle formed by (0,0),(c,0),(0,b) is right angled. So from Pythagoras, c^2+b^2=a^2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Exclamation

    Hmm

    For (iv), I've got M to be (a, b((1-cost)/sint))

    BUT when I plug that into the distance formula to get MP, it gets really long and I can find no way to simplifiy it at all!

    Any hints?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by xwrathbringerx View Post
    For (iv), I've got M to be (a, b((1-cost)/sint))

    BUT when I plug that into the distance formula to get MP, it gets really long and I can find no way to simplifiy it at all!
    I agree with that, and I think that (iv) is just plain wrong. In fact, you can easily see that it's wrong by taking t = \pi/2. Then P is the point (0,b), the tangent at P is horizontal and so M = (a,b). So in that case MP = a. The normal at P is then the y-axis, so G is at the origin and SG = ae. Therefore SG = eMP (for that particular point), so that SG \ne e^2MP.

    For a general point P, the ratio SG/MP will not be constant. The simplest formulas I can get are SG = ae(1-e\cos t) and MP = \frac{a(1-\cos t)\sqrt{1-e^2\cos^2t}}{\sin t}, from which you can see that the ratio SG/MP depends on t.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More ellipse problems
    Posted in the Geometry Forum
    Replies: 11
    Last Post: May 12th 2010, 03:51 AM
  2. locus and ellipse problems
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 10th 2009, 06:34 AM
  3. problems with ellipse
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 8th 2009, 12:08 PM
  4. Ellipse
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 1st 2008, 06:57 AM
  5. ellipse
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 30th 2008, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum