question_answer

Minimum area of the triangle by any tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] with the coordinate axes is [IIT Screening 2005]

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

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

C)            ab   

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

Correct Answer: C

Solution :

           Equation of tangent at \[(a\cos \theta ,\ b\sin \theta )\] is                   \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\]                    \[P=\left( \frac{a}{\cos \theta },\,0 \right)\]                    \[Q=\left( 0,\,\frac{b}{\sin \theta } \right)\]                    Area of \[OPQ=\frac{1}{2}\left| \,\left( \frac{a}{\cos \theta } \right)\,\left( \frac{b}{\sin \theta } \right)\, \right|=\frac{ab}{|\sin 2\theta |}\]                    \ \[{{(\text{Area})}_{\min }}=ab\].