From a point P, perpendiculars PM and PN are drawn to x and y axes, respectively. If MN passes through fixed point (a,b), then locus of P is  

Let $\mathrm{P}\left(\mathrm{h},\mathrm{k}\right)$

So, points M and N are $\left(\mathrm{h},0\right)$ and $\left(0,\mathrm{k}\right)$, respectively.

$\mathrm{MN}$ passes through the point $\mathrm{Q}\left(\mathrm{a},\mathrm{b}\right)$

$\therefore$So, M,N and Q are collinear.

$\therefore$ $-\frac{\mathrm{k}}{\mathrm{h}}=\frac{\mathrm{b}-0}{\mathrm{a}-\mathrm{h}}$

$\therefore \mathrm{ak}-\mathrm{hk}=-\mathrm{bh}$

so locus is $\mathrm{bx}+\mathrm{ay}=\mathrm{xy}$