If P and Q are points with eccentric angles θ and θ + π 6 on the ellipse x 2 16 + y 2 4 = 1 , then the area (in sq. units) of the triangle O P Q (where O is the origin) is equal to

Step-by-step solution

Let, P = 4 cos ⁡ θ , 2 sin ⁡ θ and Q = 4 c o s θ + π 6 , 2 s i n θ + π 6 ⇒ A r e a o f Δ O P Q = 1 2 4 cos ⁡ θ 2 sin ⁡ θ 1 4 c o s θ + π 6 2 s i n θ + π 6 1 0 0 1 = 4 cos ⁡ θ s i n θ + π 6 - c o s θ + π 6 sin ⁡ θ = 4 s i n θ + π 6 - θ = 4 s i n π 6 = 2 sq. units