If the curve xy=R2−16 represents a rectangular hyperbola whose branches lie only in the quadrant in which abscissa and ordinate are opposite in sign, then
A
|R|<4
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B
|R|≥4
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C
|R|=4
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D
|R|=5
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Solution
The correct option is A|R|<4 Given xy=R2−16 any point on this curve will be (x1,y1) ∵x1,y1 are opposite in sign ∴x1y1<0⇒R2−16<0⇒|R|<4