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Question

The normal to the rectangular hyperbola xy=c2 at the point 't1' meets the curve again at the point 't2'. Then the value of t31t2is


A

1

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B

-1

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C

c

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D

-c

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Solution

The correct option is B

-1


Normal at t1 meets the curve agiain at t2.
So, the normal passes through (ct1,ct1) and (ct2,ct2)
slope of normal = ct2ct1ct2ct1=1t1t2
Also equation of normal at 't1' is
xt31yt1ct41+c=0
Slope of normal = t21=1t1t2
t31t2=1


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