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Question

The normal to curve xy=4 at the point (1,4) meets the curve again at point

A
(4,1)
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B
(8,12)
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C
(16,14)
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D
(1,4)
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Solution

The correct option is C (16,14)
Let the point P be (2t1,2t1)
Slope of the normal through P is t21(1)
Let the normal at P meet again at Q(2t2,2t2)
Slope of the normal at QP=2/t22/t12t22t1=1t1t2(2)
From (1) and (2),
t31.t2=1

Given point is P(1,4)
t1=1/2;t2=8
Thus Q will be (16,14)

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