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Question

If P(x1,y1),Q(x2,y2),R(x3,y3) and S(x4,y4) are four concyclic points on the rectangular hyperbola xy=c2, then coordinates of the orthocenter of the PQR is

A
(x4,y4)
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B
(x4,y4)
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C
(x4,y4)
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D
(x4,y4)
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Solution

The correct option is C (x4,y4)
Orthocentre of triangle formed by taking three points on rectangular hyperbola lies on same hyperbola.
let (xi,yi)=(cti,cti) where i=1,2,3,4 represents given points

If (cti,cti), where i=1,2,3 are the vertices of the triangle then the orthocentre is (ct1t2t3,ct1t2t3),

Now let circle equation be
x2+y2+2gx+2fy+r=0x2+c4x2+2gx+2fc2x+r=0x4+2gx3+rx2+2fc2x+c4=0
Absissa of intersection points are roots of above equation.
So, using product of roots
x1x2x3x4=c4
t1t2t3t4=1
Hence, orthocentre is (ct4,ct4)=(x4,y4).

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