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Question

If normal to the rectangular hyperbola xy=4 at the point t1(=4) meets the curve again at the point t2, then 1|t2|=

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Solution

In the rectangular hyperbola xy=4, equation of normal at t1 is
(y2t1)=t21(x2t1)y12=16(x8)(1)
Here (1) passes through (2t2,2t2)
(2t212)=16(2t28)
4t22t2=32(t24)
t2=164 [t1t2]

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