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Math Help - coordinate ellipse

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    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
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  2. #2
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    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.
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    how have yu suppose the ordinate of centre of ellipse as 1
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  4. #4
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    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).
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    Senior Member pankaj's Avatar
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    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
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