question_answer

The area of parallelogram formed by the tangents at the ends of conjugate diameters for the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{4}=1\] is

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 units