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An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( {\frac{1}{2},\;1} \right)$. 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
$\frac{{{{(x - 1/3)}^2}}}{{1/9}} + \frac{{{{(y - 1)}^2}}}{{1/12}} = 1$
$\frac{{{{(x - 1/3)}^2}}}{{1/9}} + \frac{{{{(y + 1)}^2}}}{{1/12}} = 1$
$\frac{{{{(x - 1/3)}^2}}}{{1/9}} - \frac{{{{(y - 1)}^2}}}{{1/12}} = 1$
$\frac{{{{(x - 1/3)}^2}}}{{1/9}} - \frac{{{{(y + 1)}^2}}}{{1/12}} = 1$
Solution
(a) 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 $x = – 1$
Out of these, $x = 1$ is nearer to the point $P(1/2,1)$.
Thus a directrix of the required ellipse is $x = 1.$
If $Q(x,y)$ is any point on the ellipse,
then its distance from the focus is $QP = \sqrt {{{\left( {x – \frac{1}{2}} \right)}^2} + {{(y – 1)}^2}} $
and its distance from the directrix $x = 1$is $|x – 1|$.
By definition of ellipse, $QP = e|x – 1|$
$ \Rightarrow \sqrt {{{\left( {x – \frac{1}{2}} \right)}^2} + {{(y – 1)}^2}} = \frac{1}{2}|x – 1|$
==> $3{x^2} – 2x + 4{y^2} – 8y + 4 = 0$
or $\frac{{{{\left( {x – \frac{1}{3}} \right)}^2}}}{{1/9}} + \frac{{{{(y – 1)}^2}}}{{1/12}} = 1$.
