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Question

Normal at (5,3) of rectangular hyperbola xyy2x2=0 intersects it again at a point

A
(34,14)
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B
(1,0)
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C
(1,1)
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D
(0,2)
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Solution

The correct option is A (34,14)
xyy2x2=0 (x1)(y2)=4
XY=4
Normal at (ct,ct) intersect it again at (ct,ct) then we know that t=1t3
x1=X=ct
2t=4
t=2
Hence t=18
The intersection point will be (X,Y)
=(14,16)
(x,y)=(34,14)

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