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Q.

If A and B are two variable points on axes of x and y respectively, such that OA+OB=c, the locus of the foot of the perpendicular from origin to AB is.

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a

x²+y²=cxy

b

x+y=cxy

c

(x+y)(x²+y²)=cxy

d

x²+y²=c²

answer is C.

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Detailed Solution

$\begin{array}{l} OP ⊥ AB, m_{AB} = - h / k \\ \vec{AB} is y - k = \frac{- h}{k} (x - h) \\ \Rightarrow \frac{x}{\frac{h^{2} + k^{2}}{h}} + \frac{y}{\frac{h^{2} + k^{2}}{k}} = 1 \\ OA + O B = c \Rightarrow \frac{h^{2} + k^{2}}{h} + \frac{h^{2} + k^{2}}{k} = c \\ \Rightarrow (h^{2} + k^{2}) (h + k) = chk \end{array}$
Locus is (x²+y²)(x+y)=cxy

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