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Question

A variable circle whose centre lies on y236=0 cuts rectangular hyperbola xy=16 at (4ti,4ti),i=1,2,3,4 then 4i=11ti can be

A
3
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B
3
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C
2
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D
2
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Solution

The correct option is B 3
Let circle be x2+y2+2gx+2fy+c=0(1)
It cuts rectangular hyperbola xy=16x=16y(2)
From equation (1) and (2), we get
256y2+y2+32gy+2fy+c=0
y4+2fy3+cy2+32gy+256=0(3)
Here, we have 4 solutions for equation (3)
We have, (xi,yi)(4ti,4ti)
yi=y1+y2+y3+y4=4(1t1+1t2+1t3+1t4)
2f=4(1t1+1t2+1t3+1t4)
f2=(1t1+1t2+1t3+1t4)
Since, centre lies on y236=0
f2=36f=±6
Hence, (1t1+1t2+1t3+1t4)=±62=±3

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