Let the equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1\qquad a > b$
Then the foci are $S(ae,0)$ and $S'(-ae,0)$
Length of minor axes =2b
Length of lactus rectum =$\large\frac{2b^2}{a}$
According to the question
$BB'=SS'$
$2b=2ae$
$b=ae$
$\large\frac{2b^2}{a}$$=10$
$b^2=5a$
$b^2=a^2(1-e^2)$
$a^2e^2=a^2(1-e^2)$
$e=\large\frac{1}{\sqrt 2}$
$b=\large\frac{a}{\sqrt 2}$
$b^2=\large\frac{a^2}{2}$
$5a=\large\frac{a^2}{2}$
$a=10$
$b^2=5\times 10=50$
Hence equation of ellipse
$\large\frac{x^2}{100}+\frac{y^2}{50}$$=1$
Hence (a) is the correct answer.