Explanation

(b)
Given equation of ellipse can be rewritten as
$\frac{(x - 2)^{2}}{25} + \frac{(y + 3)^{2}}{16} = 1 \Rightarrow \ \frac{X^{2}}{25} + \frac{Y^{2}}{16} = 1$
Where $X = x - 2,\ Y = y + 3$
Here, $a > b$
$\therefore\ \ e = \sqrt{1 - \frac{b^{2}}{a^{2}} =}\sqrt{1 - \frac{16}{25}} = \frac{3}{5}$
$\therefore Focus\ ( \pm ae,\ 0) = ( \pm 3,0)$
$\Rightarrow \ \ x - 2 = \pm 3,\ y + 3 = 0$
$\Rightarrow \ \ x = 5,\ = - 1,\ y = - 3$
$\therefore Foci\ are\ ( - 1, - 3)and\ ( - 1,\ - 3)$
Distance between (2,-3) and (-1,-3)
$= \ \sqrt{(2 + 1)^{2} + ( - 3 + 3)^{2}} = 3$
and distance between $(2,\ - 3)$ and $(5,\ - 3)$
$= \sqrt{(2 - 5)^{2} + ( - 3 + 3)^{2}} = 3$
Hence, sum of the distance of point $(2,\ - 3)$from the foci
$= \ 3 + 3 = 6$