Question

$A$ and $B$ are two variable points on $x,y$ axes respectively such that $OA+OB=C$. The locus of foot of perpendicular from origin on this line is

• A. $\left(x^2+y^2\right)\left(x-y\right)=cxy$
• B. $x^2+y^2=2cxy$
• C. $(x^2+y^2)(x+y)=2cxy$
• D. $(x+y)(x^2+y^2)=cxy$

Answer 1

Answer: (A)

$\left(x^2+y^2\right)\left(x-y\right)=cxy$

$LettheequationoflineABbemx+axintercept=\frac{-a}{m}abd,yintercept-aGiventhata+\left|\frac{-a}{m}\right|=c$

$Letthefootofperpendicularbe\left(h,k\right)\therefore \frac{c_m}{m+1}=\frac{-\frac{c_n}{n}}{1-\frac{n}{k}}=\frac{ch}{h-k}k=mn+ak=\frac{-h^2}{k}+\frac{ch}{h-k}y=\frac{-x^2}{y}+\frac{cx}{x-y}\Rightarrow \left(x^2+y^2\right)\left(x-y\right)=cxy$

$Hence,optionAisthecorrectanswer.$