Any tangent at a point $P (x,y)$ to the ellipse $x^2/8 + y^2/18 =1$ meets the coordinate axes in the points $A$ and $B$ such that the the area of triangle $OAB$ is least where $O$ is the origin. Then point $P$ is of the form $(m,\,n)$ where $m$ and $n$ is ?
I tried using the parametric coordinates of the ellipse and tried to form the equation of the lines to determine the points $A$ and $B$ but it didn't help much.
Let $E$ be an ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. $$ As you know (I suppose), tangent equation at point $P(x_0,\, y_0)$ is $$ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1. $$ Points of intersection with $x$-axis and $y$-axis are $$ A(a^2/x_0,\, 0),\quad B(0,\, b^2/y_0) $$ Area of triangle $\triangle OAB$ is $$ S = \frac12\times\frac{a^2}{x_0}\times\frac{b^2}{y_0}. $$ We should to find a minumum of this. But $(x_0,\,y_0)\in E$; so, $$ x_0 = a\cos t_0,\enspace y_0 = b\sin t_0, $$ and $$ S = \frac{ba}{\sin(2t_0)}. $$ Minumum $S$ is at $$\sin(2t_0)=1\Longrightarrow t_0 = \frac{\pi}{4}$$ (we work in the first quadrant by symmetry). So, $$ x_0 = \frac{a}{\sqrt2},\enspace y_0 = \frac{b}{\sqrt2}. $$
In your case and notation $$ m=2,\enspace n = 3. $$
The tangent line at $(x_0,y_0)$ is $$ \frac{x_0x}{8}+\frac{y_0 y}{18}=1. $$ Now set $x=0$ to get $B=(0,18/y_0)$ and set $y=0$ to get $A=(8/x_0,0)$. The area is $$ \frac{18}{y_0} \cdot \frac{8}{x_0} \cdot \frac{1}{2} = \frac{18}{x_0 y_0}. $$ Recall that $(x_0,y_0)$ lies on the ellipse, so that you must compute $$ \min \left\{ \frac{18}{x_0 y_0} \mid \frac{x_0^2}{8}+\frac{y_0^2}{18}=1 \right\}. $$ You can now use Lagrange multipliers, or express $y_0$ in terms of $x_0$. Or, even better, use the polar representation $x=\sqrt{8}\cos \theta$, $y=\sqrt{18}\sin\theta$.
