The minimum area of a triangle formed by any tangent to the ellipse...

Let $(h, k)$ be the point on ellipse through which tangent is passing. Equation of tangent at (h, k)=$\frac{x h}{16}+\frac{y k}{81}=$1 at $y=0, x=\frac{16}{h}$ at $x=0, y=\frac{81}{k}$ Area of \mathrm{AOB}=\frac{1}{2} \times\left(\frac{16}{h}\right) \times\left(\frac{81}{k}\right)=\frac{648}{h k}\mathrm{A}^2=$\frac{(648)^2}{h^2 k^2}(h, k)$ must satisfy equation of ellipse $$\begin{aligned}&\frac{h^2}{16}+\frac{k^2}{81}=1 \\ &h^2=\frac{16}{81}\left(81-k²\right) \end{aligned}$$Putting value of $h^2$ in equation (1) \mathrm{A}^2=$\frac{81(648)^2}{16 \times k^2\left(81-k^2\right)}=\frac{\alpha}{81 k^2-k^4}$differentiating w.r. to $k$$\begin{aligned}&2 \mathrm{AA}^{\prime}=\alpha\left(\frac{-1}{81 k^2-k^4}\right)\left(162 k-4 k³\right) \\ &2 \mathrm{AA}^{\prime}=-2 \mathrm{~A}\left(81 k-4 k³\right) \\ &\Rightarrow \mathrm{A}^{\prime}=-81 k-4 k³ \\ &\text{ Put } \mathrm{A}^{\prime}=0 \\ &\Rightarrow 162 k-4 k³=0, k\left(162-4 k²\right)=0 \\ &\Rightarrow k=0, k=\pm \frac{9}{\sqrt{2}}\\ &\mathrm{~A}^{\prime \prime}=-\left(81-12 k²\right) \end{aligned}$$For both value of k, A"$=405 \gt 0$Area will be minimum for$k=\pm \frac{9}{\sqrt{2}}$$\begin{aligned}&h^2=\frac{16}{81}\left(81-k²\right)=8 \\ &h=\pm2 \sqrt{2} \end{aligned}$$$Area of triangle$\mathrm{AOB}=\frac{648 \times \sqrt{2}}{2 \sqrt{2} \times 9}=36$ sq unit