The minimum area of triangle formed by the tangent to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` and coordinate axes is `a b` sq. units (b) `(a^2+b^2)/2` sq. units `((a+b)^2)/2` sq. units (d) `(a^2+a b+b^2)/3` sq. units

Options

A. ab sq. units
B. `(a^(2)+b^(2))/(2)` sq. units
C. `((a+b)^(2))/(2)` sq. units
D. `(a^(2)+ab+b^(2))/(3)` sq. units

Explanation

Tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
at `p( a cos theta, b sin theta)` is given by
`(xcos theta)/(a)+(y sin theta)/(b)=1`
It meets he coordinate axes at `A(a sec theta, 0) and B(0, b "coses" beta)`
or `Delta=(ab)/(sin 2 theta)`
For area to be minimum, `sin theta` should be maxiumum and we know that teh maxiumum value of `sin theta` is 1. Therefore,
`Delta_("max")=ab`