question_answer

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

A) \[ab\]sq. units

B) \[\frac{{{a}^{2}}+{{b}^{2}}}{2}sq.units\]

C) \[\frac{{{\left( a+b \right)}^{2}}}{2}\operatorname{sq}.units\]

D) \[\frac{{{a}^{2}}+ab+{{b}^{2}}}{3}sq.units\] 

Correct Answer: A

Solution :

Any tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at \[P\left( a\cos \theta ,b\sin \theta  \right)is\,\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\]

It meets co-ordinate axes at \[A\left( a\sec \theta ,0 \right)\]and \[B\left( 0,b\cos c\theta  \right)\]

\[\therefore \]Area of \[\Delta OPB=\frac{1}{2}\times a\sec \theta \times b\cos \theta \]\[\Rightarrow \Delta =\frac{ab}{\sin 2\theta }\]

For \[\Delta \]to be min, sin 2\[\theta \]should be max. and we know max value of sin=1

\[\therefore {{\Delta }_{\max }}=ab\,\operatorname{sq}.units\].