Question

An ellipse has eccentricity $\frac{1}{2}$ and one focus is at the point $P(\frac{1}{2},1)$. If the common tangent to the circle $x^2+y^2=1$ and hyperbola $x^2-y^2=1$ which is nearer to point $P$ is directrix of the given ellipse, then the co-ordinates of centre of ellipse are

• A. $(\frac{1}{3},\frac{1}{3})$
• B. $(\frac{2}{3},1)$
• C. $(\frac{1}{3},1)$
• D. $(1,\frac{1}{3})$

Answer 1

The correct option is C $(\frac{1}{3},1)$

Given : Circle is $x^2+y^2=1$ and hyperbola is $x^2-y^2=1$
Therefore, the common tangents are
$x=\pm 1$


But $x=1$ is nearer to the point $P(\frac{1}{2},1).$
$\therefore$ Directrix of the required ellipse is $x=1$
As one of the focus is $P(\frac{1}{2},1)$, so the centre is
$C=(\frac{1}{2}-ae,1)=(\frac{1}{2}-\frac{a}{2},1)$
Distance from $C$ to directrix $=\frac{a}{e}=2a$, so
$C=(1-2a,1)$

Therefore,
$1-2a=\frac{1-a}{2}\Rightarrow a=\frac{1}{3}\therefore C=(\frac{1}{3},1)$