Results 1 to 5 of 5

Thread: coordinate ellipse

  1. #1
    Senior Member
    Joined
    Feb 2010
    Posts
    460
    Thanks
    34

    coordinate ellipse

    an ellipse has eccentricity 1/2 and a focus at the point p(1/2,1).One of its directrix is the common tangent near to the point P to the circle x^2+y^2=1 and hyperbola x^2-y^2=1 the equation of ellipse is

    i have drawn figure but unable to proceed further
    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
    Quote Originally Posted by prasum View Post
    an ellipse has eccentricity 1/2 and a focus at the point p(1/2,1).One of its directrix is the common tangent near to the point P to the circle x^2+y^2=1 and hyperbola x^2-y^2=1 the equation of ellipse is

    i have drawn figure but unable to proceed further
    The figure that you have drawn should show that the common tangent to the circle and the hyperbola is the vertical line x=1. That tells you that the ellipse has its centre on the x-axis, and its major axis parallel to the x-axis. Say that the centre is at the point (c,1). If the eccentricity is e, the major semi-axis has length a and the minor semi-axis has length b, then you should know that the distance from the centre to the focus is ae, the distance from the centre to the directrix is a/e, and b is given by b^2 = a^2(1-e^2). Use those facts to find a, b and c. Then the equation of the ellipse is \frac{(x-c)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2010
    Posts
    460
    Thanks
    34
    how have yu suppose the ordinate of centre of ellipse as 1
    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 prasum View Post
    how have yu suppose the ordinate of centre of ellipse as 1
    You know that the directrix is vertical, so that the major axis is horizontal. Therefore the y-coordinate of the centre is the same as the y-coordinate of the focus, which is given as (1/2,1).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    You need to have a sound background to answer this one.

    Equation of tangent to the given circle is y=mx+\sqrt{1+m^2} and that of the given hyperbola is y=m_{1}x+\sqrt{m_{1}^2-1}.

    If they both constitute the same line them m=m_{1} and thus

    \sqrt{1+m^2}=\sqrt{m^2-1}

    1+m^2=m^2-1

    \frac{1}{m^2}+1=1-\frac{1}{m^2}

    \frac{2}{m^2}=0

    m=\infty

    The common tangent will be a line parallel to y-axis and such a line can only be x=\pm1

    Since we are interested in the directrix nearer to the point P\big(\frac{1}{2},1\big) the equation of the directrix will be x-1=0.

    The general equation of any conic is
    (x-x_{1})^2+(y-y_{1})^2=e^2\big\{\frac{lx+my+n}{\sqrt{l^2+m^2}}\b  ig\}^2

    where (x_{1},y_{1}) is the focus of the conic, e the eccentricity and lx+my+n=0 the directrix of the conic

    Thus the required equation is

    (x-\frac{1}{2})^2+(y-1)^2=\frac{1}{4}\big\{\frac{x-1}{\sqrt{1^2+0^2}}\big\}^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. coordinate
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 12th 2010, 08:42 AM
  2. Replies: 5
    Last Post: Aug 11th 2010, 12:15 AM
  3. coordinate
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Aug 5th 2010, 01:15 PM
  4. value of x coordinate
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 23rd 2009, 08:40 AM
  5. Replies: 4
    Last Post: Aug 14th 2009, 11:48 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum