Question

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P(\frac{1}{2},1)$. Its one directrix is the common tangent, nearer to the point $P$, to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1$. The equation of the ellipse, in the standard form is..............

Answer 1

There are two common tangents to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1$. These are $x=-1$ and $x=-1$.
But $x=1$ is nearer to the point $P(1/2,1)$. Therefore, directrix of the required ellipse is $x=1$
Now, if $Q(x,y)$ is any point on the ellipse, then its distance from the focus is
$QP=\sqrt{(x-1/2{)}^2+(y-1{)}^2}$
and its distance from the directrix is $|x-1|$. By definition of ellipse,

$QP=e|x-1|\sqrt{(x-\frac{1}{2}{)}^2+(y-1{)}^2}=\frac{1}{2}|x-1|$
$\Rightarrow (x-\frac{1}{2}{)}^2+(y-1{)}^2=\frac{1}{4}(x-1{)}^2$
$\Rightarrow x^2-x+\frac{1}{4}+y^2-2y+1=\frac{1}{4}(x^2-2x+1)$
$\Rightarrow 4x^2-4x+1+4y^2-8y+4=x^2-2x+1$
$\Rightarrow 3x^2-2x+4y^2-8y+4=0$
$\Rightarrow 3\left[{\left(x-\frac{1}{3}\right)}^2-\frac{1}{9}\right]+4(y-1{)}^2=0$
$\Rightarrow 3{\left(x-\frac{1}{3}\right)}^2+4(y-1{)}^2=\frac{1}{3}$
$\Rightarrow \frac{{\left(x-\frac{1}{3}\right)}^2}{1/9}+\frac{(y-1{)}^2}{1/12}=1$