Q.

P  and Q  are two variable points on the axes of x  and y  respectively such that |OP|+|OQ|=α , then the locus of foot of perpendicular from (0, 0)  on PQ  is

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a

(x+y)(x2+y2)=a(x−y)

b

(x−y)(x2−y2)=axy

c

(x+y)(x2+y2)=axy

d

(x+y)(x2−y2)=axy

answer is C.

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Detailed Solution

Let PQ¯  be xα+yβ=1  equation of the circle passing through O,  P,  R  is x2+y2−αx=0.⇒α=x2+y2x where Ris foot of perpendicularSimilarly β=x2+y2y  ∴   |α|+|β|=aNow eliminate α, β
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