The focus and corresponding directrix of an ellipse are (3,4) and x...

Let, the lengths of the semi-major axis and semi-minor axis of the& ellipse are& $a$ & $b$ respectively, then the distance of the focus \left(3,4\right)$from the corresponding directrix \left(x + y - 1 = 0\right)$ is equal to $\dfrac{a}{e} - ae$
$⇒ \dfrac{a}{\dfrac{1}{2}} - a\left(\dfrac{1}{2}\right) = \left|\dfrac{3 + 4 - 1}{\sqrt{2}}\right| = 3\sqrt{2}$
$⇒ \dfrac{3a}{2} = 3\sqrt{2} ⇒ a = 2\sqrt{2}$
Distance between the focus and centre $= ae = \sqrt{2}$
The slope of the axis of the ellipse is $1 = tan \theta$
⇒ \left( cos\theta,$sin$\theta\right) = \left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right)$or \left(- \dfrac{1}{\sqrt{2}}, - \dfrac{1}{\sqrt{2}}\right)$
The points on the axis of the ellipse at a distance \sqrt{2} units from \left(3,4\right)are \left(3\pm \sqrt{2}\left(\dfrac{1}{\sqrt{2}}\right),4 \pm \sqrt{2}\left(\dfrac{1}{\sqrt{2}}\right)\right)<br />⇒$Points are \left(2,3\right)$ or \left(4,5\right)$, but the centre is \left(4,5\right)$ because it is far from the directrix as compared to \left(2,3\right).