If P and Q are points with eccentric angles θ and (θ + (π /6)) on the ellipse (x2/16)+(y2/4)=1, then the area (in sq. units) of the triangle OPQ (where O is the origin) is equal to

Question

If P and Q are points with eccentric angles θ and (θ + (π /6)) on the ellipse (x2/16)+(y2/4)=1, then the area (in sq. units) of the triangle OPQ (where O is the origin) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let, P=(4 cos θ, 2 sin θ) and Q=(4 cos (θ+(π/6)), 2 sin (θ+(π/6))) ⇒ AreaofΔ OPQ=(1/2)| 4cosθ 2sinθ 1 4cos(θ + (π /6)) 2sin(θ + (π /6)) 1 0 0 1 | =4(cos θ s i n (θ + (π /6)) - c o s (θ + (π /6)) sin θ ) =4sin(θ + (π /6) - θ )=4sin(π /6)=2 sq. units

Q. If P and Q are points with eccentric angles θ and (θ+6π​) on the ellipse 16x2​+4y2​=1, then the area (in sq. units) of the triangle OPQ (where O is the origin) is equal to

Let, P=(4cosθ,2sinθ) and Q=(4cos(θ+6π​),2sin(θ+6π​)) ⇒AreaofΔOPQ=21​∣∣​4cosθ4cos(θ+6π​)0​2sinθ2sin(θ+6π​)0​111​∣∣​ =4(cosθsin(θ+6π​)−cos(θ+6π​)sinθ) =4sin(θ+6π​−θ)=4sin6π​=2 sq. units

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