A) 16 sq units
B) 8 sq units
C) 32 sq units
D) \[18\sqrt{3}\] sq units
Correct Answer: C
Solution :
Area of parallelogram \[{{T}_{1}}{{T}_{2}}{{T}_{3}}{{T}_{4}}\] = 4 (Area of parallelogram \[CP\,\,{{T}_{2}}\,D\] = 4 (2 Area of \[\Delta CPD\]) = 8 (Area of \[\Delta CPD\]) \[=8\times \frac{1}{2}\,\left| \begin{matrix} 0 & 0 & 1 \\ a\,\cos \,\theta & b\,\sin \,\theta & 1 \\ -a\,\sin \,\theta & b\,\cos \,\theta & 1 \\ \end{matrix} \right|\] \[=4\,(ab\,{{\cos }^{2}}\theta +ab\,{{\sin }^{2}}\theta )\] \[=4ab=2a\times 2b\] = Product of the axes of the ellipse \[\therefore \] Area \[=4\times 4\times 2=32\] sq unitsYou need to login to perform this action.
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