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 $\overleftrightarrow{AB}is$

Let Equation of line

$\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1--\left(1\right)\\ GivenOA+OB=C\\ \Rightarrow a+b=c\\ Let\left(hk\right)bethefootof\perp rfrom\left(0,0\right)to\left(1\right)\\ \Rightarrow \frac{h-0}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}=\frac{k-0}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\right)}=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}\\ \Rightarrow ha=kb=\frac{a^2{b}^2}{a^2+b^2}\\ h=\frac{a{b}^2}{a^2+b^2},k=\frac{a^{2}b}{a^2+b^2}\end{array}$

$\begin{array}{l}h+k=\frac{ab\left(a+b\right)}{a^2+b^2},h^2+k^2=\frac{a^2{b}^4+a^4{b}^2}{{\left(a^2+b^2\right)}^2}2\Rightarrow \frac{a^2{b}^2\left(a^2+b^2\right)}{{\left(a^2+b^2\right)}^2}\Rightarrow \frac{a^2{b}^2}{a^2+b^2}\\ \left(h+k\right)\left(h^2+k^2\right)=\frac{ab\left(a+b\right)}{a^2+b^2}\times \frac{a^2{b}^2}{a^2+b^2}\\ \Rightarrow \left(a+b\right)\left(\frac{a{b}^2}{a^2+b^2}.\frac{a^{2}b}{a^2+b^2}\right)\\ \left(h+k\right)\left(h^2+k^2\right)=chk\\ \therefore locusof\left(hk\right)is\left(x+y\right)\left(x^2+y^2\right)=xyc\end{array}$