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

Any tangent to the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\text{ at }P(a\cos \theta ,b\sin \theta )\text{ is }$

given by

$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

$\text{ It meets the coordinate axes at }A(a\sec \theta ,0)\text{ and }B(0,b\mathrm{cosec}\theta )\text{ . Therefore, }$

$\text{ Area of }\mathrm{\Delta }OAB=\frac{1}{2}\times a\sec \theta \times b\mathrm{cosec}\theta$

$\text{ or }\mathrm{\Delta }=\frac{ab}{\sin 2\theta }$

$\text{ For area to be minimum, }\sin 2\theta \text{ should be maximum and we }$

$\text{ know that the maximum value of }\sin 2\theta \text{ is }1\text{ . Therefore, }$

${\mathrm{\Delta }}_{max}=ab$