If the normal to the rectangular hyperbola xy=4 at the point P(4) meets the curve again at the point Q(t2), then find the value of 1|t2|
A
64.0
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B
64.00
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C
64
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Solution
If the normal to the rectangular hyperbola xy=c2
at the point (t1) meets the curve again at the point (t2), then ⇒t31t2=−1
Normal is drawn at Point P(4) ∴t1=4
So, substitute t1=4 in t31t2=−1 ⇒43t2=−1 ⇒t2=−164 ⇒1|t2|=64