Question

An ellipse has eccentricity $\frac {1}{2}$ and one focus at the point $P \bigg(\frac{1}{2}, 1 \bigg)$. 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 $
$\Rightarrow x^2 - x + \frac{1}{4} + y^2 - 2y + 1 =\frac {1}{4} (x^2 - 2x + 1)$
$\Rightarrow 4x^2 - 4x + 1+ 4y^2- 8y + 4 = x^2 - 2x + 1$
$\Rightarrow 3 x^{2}-2 x+4 y^{2}-8 y+4=0$
$\Rightarrow 3\left[\left(x-\frac{1}{3}\right)^{2}-\frac{1}{9}\right]+4(y-1)^{2}=0$
$\Rightarrow 3\left(x-\frac{1}{3}\right)^{2}+4(y-1)^{2}=\frac{1}{3}$
$\Rightarrow \frac{\left(x-\frac{1}{3}\right)^{2}}{1 / 9}+\frac{(y-1)^{2}}{1 / 12}=1$