A and B are two variable points on $x, y$ axes respectively such that $OA + OB = c$. The locus of the foot of the perpendicular from origin on this line $\overline{AB}$ is

Let P(h, k) be the foot of the perpendicular from the origin O(0, 0). Slope of $\overrightarrow{OP}\text{ is }\frac{k}{h}\text{. }$

$\therefore$Slope of AB $\overleftrightarrow{AB}\text{ is }-\frac{h}{k}.$

Equation to $\overrightarrow{AB}\text{ is }y-k=-\frac{h}{k}(x-h)$

$\Rightarrow hx+ky=h^2+k^2$

$\therefore A\left(\frac{h^2+k^2}{h},0\right),B\left(0,\frac{h^2+k^2}{k}\right)$

$OA+OB=c\Rightarrow \frac{h^2+k^2}{h}+\frac{h^2+k^2}{k}=c$