If normal to the rectangular hyperbola xy=4 at the point t1(=4) meets the curve again at the point t2, then 1|t2|=
A
64.0
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B
64
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C
64.00
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Solution
In the rectangular hyperbola xy=4, equation of normal at t1 is (y−2t1)=t21(x−2t1)⇒y−12=16(x−8)⋯(1)
Here (1) passes through (2t2,2t2) ⇒(2t2−12)=16(2t2−8) ⇒4−t22t2=32(t2−4) ⇒t2=−164[∵t1≠t2]