Any point on the hyperbola $x^2-y^2=1$ is (sec $\theta$, tan $\theta$).
Then tangent at (sec $\theta$, tan $\theta$) to $x^2-y^2=1$ is
 $xsec\theta -y\text{tan}\theta =1 .........(1)$
 this will also be a tangent to  $x^2+y^2=1$ if radius = length of perpendicular from (0,0) to the equation (1)
or $1=\frac{1}{\sqrt{{sec}^2\theta +{\text{tan}}^2\theta }}$
 or ${sec}^2\theta +{\text{tan}}^2\theta =1$
 or  $2{tan}^2\theta =0$
or   $tan\theta =0 \therefore \theta =0\text{,} \pi$
 $\therefore$   putting for θ in (1), the common tangents are $\pm x=1$,
 $x=1\&x+1=0$
 $x=1$ is nearer to $F\left(\frac{1}{2},1\right)$ than $x+1=0$
 $\therefore$ the ellipse has the following focus $=\left(\frac{1}{2},1\right)$
 corresponding directrix is x -1 = 0 and $e=\frac{1}{2}$
 $\therefore$ by focus - directirx property, the equation of the ellipse is 
 $\sqrt{{\left(\text{x}-\frac{1}{2}\right)}^2+{\left(\text{y}-1\right)}^2}=\frac{1}{2}\left(\frac{\text{x}-1}{\sqrt{1^2+0^2}}\right)$
or  3x² + 4y² - 2x - 8y + 4 = 0