Q.

If the normals drawn to the hyperbola xy=4 at ( αi,βi ) (i = 1, 2, 3, 4) are concurrent at the point (a, b) then α1+α2+α3+α4β1+β2+β3+β4α1α2α3α4=
 

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a

-16ba

b

-16ab

c

4ab

d

4ba

answer is B.

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Detailed Solution

Given hyperbola xy=4

let (2t,2t) be any point on given curve

Now equation of normal at 't' is y-ct=t2(x-ct)

If it passes through (a, b) then

b-ct=t2(a-ct) bt-c=t3(a-ct) bt-c=at3-ct4 ct4-at3+bt-c=0 2t4-at3+bt-2=0 (where c2=4  c=2) let t1, t2, t3, t4  be  four roots. t1+t2+t3+t4=a2,   t1t2=O,   t1t2t3=-b2, and t1t2t3t4=-1.Now α1+α2+α3+α4=(2t1+2t2+2t3+2t4)                                       =2a2=a          α1.α2.α3.α4=(2t1) (2t2) (2t3) (2t4)=16(-1)=-16 Now β1+β2+β3+β4=2t1+2t2+2t3+2t4                                       =21t1+1t2+1t3+1t4=2  t1t2t3t1t2t3t4                                        =2(-b2)-1=b  Req. value =-16ab 

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