Let S and S^ ′ are the focii of the ellipse x = 3 + 5cos θ, y = - 2...

Assume the base of $\text{\Delta}SS^{′} B$ is $SS^{′}$, then
base $= SS^{′} = 2ae$
height $= \dfrac{1}{2}\left(\dfrac{2b^{2}}{a}\right) = \dfrac{lengthoflatusrectum}{2}$
$⇒ \text{area of \Delta}BSS^{′} = \dfrac{1}{2} \left(2ae\right)\left(\dfrac{b^{2}}{a}\right)= b^{2}e$
Now, equation of the ellipse is $\dfrac{{\left(x - 3\right)}^{2}}{25} + \dfrac{{\left(y + 2\right)}^{2}}{16} = 1$
$⇒ a^{2} = 25, b^{2} = 16, e^{2} = \dfrac{9}{25}$
$⇒ b^{2}e = 16 \times \dfrac{3}{5} = \dfrac{48}{5}$

Hence, area of $\text{\Delta}BSS^{′}= \dfrac{48}{5}$& sq. units