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The sum of the distance of any point on the ellipse $ 4x^{2}+9y^{2}=36 $ from $(\sqrt{5},0) and (-\sqrt{5} ,0) $ is

 \[\begin{array}{1 1}(1)4&(2)8\\(3)6&(4)18\end{array}\]

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$4x^2+9y^2=36$
$\large\frac{4x^2}{36}+\frac{9y^2}{36}$$=1$
$\large\frac{x^2}{9} +\frac{y^2}{4}$$=1$
$a^2=9$
$b^2=4$
$e = \sqrt {1- \large\frac{b^2}{a^2}}$
$\qquad= \large\frac{\sqrt 5}{3}$
$S(ae,0) = \bigg(3 \times \large\frac{\sqrt 5}{3},0 \bigg)= (\sqrt 5,0)$
$S'(-ae,0) = \bigg(-3 \times \large\frac{\sqrt 5}{3},0 \bigg)= (-\sqrt 5,0)$
$\therefore (\sqrt 5,0)$ & $(- \sqrt 5 ,0)$ are the foci of the ellipse
Let P be the point o the ellipse
$4x^2+9y^2 =36$ whose foci are $ S ( \sqrt 5,0)$ and $S' ( - \sqrt 5 ,0)$ then
$SP+S'P =2a$
$2 \times 3=6$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
 

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