1. ## 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

2. Originally Posted by prasum
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 $\frac{(x-c)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1.$

3. how have yu suppose the ordinate of centre of ellipse as 1

4. Originally Posted by prasum
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).

5. 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

$(x-\frac{1}{2})^2+(y-1)^2=\frac{1}{4}\big\{\frac{x-1}{\sqrt{1^2+0^2}}\big\}^2$