Question

Let $P$ be a point in the first quadrant lying on the ellipse $\frac{x^2}{8}+\frac{y^2}{18}=1.$ Let $AB$ 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 $\Delta OAB$ is (in square units)

Answer 1

Answer: 12

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 $(0,$, $\sqrt{18}\mathrm{cosec}\theta )$ respectively.
Area of $\Delta OAB=\frac{1}{2}\times \sqrt{8}\sec \theta \times \sqrt{18}\mathrm{cosec}\theta =\frac{12}{\sin 2\theta }$
$\therefore$ minimum area $=12$ square units