Question

Let $P$ be a point in the first quadrant lying on the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{18}=1 .$ Let $A B$ be the tangent at $P$ to the ellipse meeting the $x$-axis at $A$ and $y$ axis at $B$. If $O$ is the origin, then the minimum possible area of $\Delta O A B$ is (in square units)

Answer 1

Any point $P$ on the ellipse will be $(\sqrt{8} \cos , \sqrt{18} \sin \theta)$. Now equation of the tangent at $P$
$\frac{x}{\sqrt{8}} \cos \theta+\frac{y}{\sqrt{18}} \sin \theta=1$
the coordinates of $A \& B$ will be $(\sqrt{8} \sec \theta, 0)$ and $(0,$, $\sqrt{18} \operatorname{cosec} \theta)$ respectively.
Area of $\Delta O A B=\frac{1}{2} \times \sqrt{8} \sec \theta \times \sqrt{18} \operatorname{cosec} \theta=\frac{12}{\sin 2 \theta}$
$\therefore $ minimum area $=12$ square units