Let, the foot of perpendiculars from $P$ on $OA$ & $OB$ are $C$ & $D$ respectively
Let, $\mathrm{\angle }PAC=\theta =\mathrm{\angle }BPD$

Now, $CA=k\cot \theta$ $\&$ $BD=h\tan \theta$
Area of $\mathrm{\Delta }OAB=$ area of $\mathrm{\Delta }PCA+$ area of $\mathrm{\Delta }BPD+$ area of the rectangle $PDOC$
$\Rightarrow$ Area of $\mathrm{\Delta }OAB=\frac{1}{2}{k}^2\cot \theta +\frac{1}{2}{h}^2\tan \theta +hk=\frac{1}{2}{\left(k\sqrt{\cot \theta }-h\sqrt{\tan \theta }\right)}^2+2hk$
$\Rightarrow$ The minimum area of $\mathrm{\Delta }OAB$ is $2hk$ when $k\sqrt{\cot \theta }=h\sqrt{\tan \theta }$$\Rightarrow \tan \theta =\frac{k}{h}$
$\Rightarrow h\sin \theta =k\cos \theta \Rightarrow h\sec \theta =k\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\theta$
$\Rightarrow PB=PA\Rightarrow OP$ is the median