If the tangent at a point on the ellipse `x^2/27+y^2/3=1` meets the coordinate axes at `A` and `B,` and the origin, then the minimum area (in sq. units) of the triangle `OAB` is:

Options

A. 9
B. `(9)/(2)`
C. `9sqrt(3)`
D. `3sqrt(3)`

Answer

A

Explanation

Let `P(3sqrt(3) cos theta, sqrt(3) sin theta)` be a point on the ellipse. Then, the equation of the tangent at P is
`(x)/(3sqrt(3)) cos theta +(y)/(sqrt(3)) sin theta =1`.
This meets the coordinate axes at `A(3sqrt(3) sec theta, 0) and B(0, sqrt(3) " cosec" theta)`.
Let `Delta` be the area of `Delta OAB.` Then,
`Delta=(1)/(2)(OAxxOB)=(1)/(2) xx (3sqrt(3))/(cos theta)xx (sqrt(3))/(sin theta)=(9)/(sin2theta)`
Clearly, `Delta` is minimum when `sin 2 theta` is maximum. The maximum value of `sin 2 theta` is 1. Hence, the minimum value of `Delta` is 9.