• # question_answer The distances of a point (7, -3) from the foci of the ellipse $3{{x}^{2}}+4{{y}^{2}}-12x+24y+36=0$are 'a' and 'b' respectively, then a and b satisfy A)  $a+b=10$                B)  $a+b=6$ C) $|b-a|\,=\,14$                 D)  $|a\pm b|\,=\,2009$

The equation of the ellipse can be written as $\frac{{{(x-2)}^{2}}}{4}+\frac{{{(y+3)}^{2}}}{3}=1,$ whose centre is (2, -3) and vertices are $A(4,\,\,-3)$ and $A'(0,\,-3)$ foci are $S(3,-3)$ and $S'(1,-3)$. $\left( \because \,\,e=\frac{1}{2} \right)$ a = Distance between P(7, -3) and S (3, -3) = 4 and b = distance between P(7, - 3) and S? (1, -3) = 6 $\therefore$    $a+b=4+6=10$