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 OPQ (where O is the origin) is equal to

Correct answer: 2 Let, P = (4cos θ,2sin θ) and Q = (4 (θ + (π)/(6)), 2 (θ + (π)/(6))) ⇒ Areaof OPQ = (1)/(2)| cases 4cos θ & 2sin θ & 1 4 (θ + (π)/(6)) & 2 (θ + (π)/(6)) & 1 0 & 0 & 1 cases | = 4(cos θ sin(θ + (π)/(6)) - (θ + (π)/(6))sin θ) = 4 (θ + (π)/(6) - θ) = 4sin(π)/(6) = 2 sq. units

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If P and Q are points with eccentric angles θ and (θ + (π)/(6)) on...

P and Q

are points with eccentric angles

\theta

and

\left(\theta + \dfrac{\pi}{6}\right)

on the ellipse

\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{4} = 1,

then the area (in sq. units) of the triangle

OPQ

(where

O

is the origin) is equal to

Correct answer

Step-by-step solution

Let,

P =

\left(4cos \theta,2sin \theta\right)

and Q = \left(4\cos \left(\theta + \dfrac{\pi}{6}\right), 2\sin \left(\theta + \dfrac{\pi}{6}\right)\right)

⇒ Areaof\delta OPQ = \dfrac{1}{2}\left|

\begin{cases}4cos \theta & 2sin \theta & 1 \\ 4\cos \left(\theta + \dfrac{\pi}{6}\right) & 2\sin \left(\theta + \dfrac{\pi}{6}\right) & 1 \\ 0 & 0 & 1\end{cases}

\right|

= 4\left(cos \theta sin\left(\theta + \dfrac{\pi}{6}\right) - \cos \left(\theta + \dfrac{\pi}{6}\right)sin \theta\right)

= 4\sin \left(\theta + \dfrac{\pi}{6} - \theta\right) = 4sin\dfrac{\pi}{6} = 2

sq. units

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