Let S and S ′ are the focii of the ellipse x = 3 + 5 cos ⁡ θ , y = - 2 + 4 sin ⁡ θ . If B is one of the ends of one of the latus rectum, then the area (in sq. units) of the triangle B S S ′ is equal to

Step-by-step solution

Assume the base of Δ S S ′ B is S S ′ , then base = S S ′ = 2 a e height = 1 2 2 b 2 a = length   of   latus   rectum 2 ⇒ area of Δ B S S ′ = 1 2 2 a e b 2 a = b 2 e Now, equation of the ellipse is x - 3 2 25 + y + 2 2 16 = 1 ⇒ a 2 = 25 , b 2 = 16 , e 2 = 9 25 ⇒ b 2 e = 16 × 3 5 = 48 5 Hence, area of Δ B S S ′ = 48 5 sq. units