question_answer

The minimum area of the triangle formed by any tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] with the coordinate axes is

A) \[{{a}^{2}}+{{b}^{2}}\]                  

B)  \[\frac{{{\left( a+b \right)}^{2}}}{2}\]

C) ab                                          

D)  \[\frac{{{\left( a-b \right)}^{2}}}{2}\]

Correct Answer: C

Solution :

Equation of tangent at \[{{X}^{2}}=3{{R}^{2}}\] to the ellipse is                            \[X=\sqrt{3}R\]          Coordinates of P and Q are \[e=M\frac{di}{dt}\] respectively. Now, area of       \[=0.005\times \frac{d}{dt}\left( {{i}_{0}}\sin \omega t \right)\] \[=0.005\times {{i}_{0}}\cos \omega t\]  Minimum area =  ab