If the tangent at a point on the ellipse $\frac{x^2}{27}+\frac{y^2}{3}=1$ meets the coordinate axes at A and B and, O is the origin, the minimum area (in sq. units). of the triangle OAB, is
Answer & explanation
Correct answer: option 1
Let $P(3 \sqrt{3} \cos \theta, \sqrt{3} \sin \theta)$ be a point on the ellipse. Then, the equation of the tangent at P is
$\frac{x}{3 \sqrt{3}} \cos \theta+\frac{y}{\sqrt{3}} \sin \theta=1$
This meets the coordinate axes at $A(3 \sqrt{3} \sec \theta, 0)$ and $B(0, \sqrt{3} ~cosec ~\theta)$.
Let Δ be the area of $\triangle O A B$. Then,
$\Delta=\frac{1}{2}(O A \times O B)=\frac{1}{2} \times \frac{3 \sqrt{3}}{\cos \theta} \times \frac{\sqrt{3}}{\sin \theta}=\frac{9}{\sin 2 \theta}$
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.