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Question: Area of a triangle whose vertices are \((a\cos\theta,b\sin\theta),\) \(( - a\sin\theta,b\cos\theta)\...

Area of a triangle whose vertices are (acos⁡θ,bsin⁡θ),(a\cos\theta,b\sin\theta), (−asin⁡θ,bcos⁡θ)( - a\sin\theta,b\cos\theta) and (−acos⁡θ,−bsin⁡θ)( - a\cos\theta, - b\sin\theta) is.

A

acos⁡θsin⁡θa\cos\theta\sin\theta

B

absin⁡θcos⁡θab\sin\theta\cos\theta

C

12ab\frac{1}{2}ab

D

abab

Answer

abab

Explanation

Solution

Area =12∣acos⁡θbsin⁡θ1−asin⁡θbcos⁡θ1−acos⁡θ−bsin⁡θ1∣= \frac { 1 } { 2 } \left| \begin{array} { c c c } a \cos \theta & b \sin \theta & 1 \\ - a \sin \theta & b \cos \theta & 1 \\ - a \cos \theta & - b \sin \theta & 1 \end{array} \right|

=12(a×b)∣cos⁡θsin⁡θ1−sin⁡θcos⁡θ1−cos⁡θ−sin⁡θ1∣= \frac { 1 } { 2 } ( a \times b ) \left| \begin{array} { c c c } \cos \theta & \sin \theta & 1 \\ - \sin \theta & \cos \theta & 1 \\ - \cos \theta & - \sin \theta & 1 \end{array} \right|

=ab2[cos⁡θ(cos⁡θ+sin⁡θ)−sin⁡θ(−sin⁡θ+cos⁡θ)= \frac { a b } { 2 } [ \cos \theta ( \cos \theta + \sin \theta ) - \sin \theta ( - \sin \theta + \cos \theta ) +1(sin⁡2θ+cos⁡2θ)]\left. + 1 \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta \right) \right] =ab2(1+1)=ab= \frac { a b } { 2 } ( 1 + 1 ) = a b.