A≡(−4,0),B≡(4,0) . M and N are the variable points of y-axis such that M lies below N and MN = 4. Line joining AM and BN intersect at 'P'. Locus of 'P' is

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2xy−16−x2=0

2xy+16−x2=0

2xy+16+x2=0

2xy−16+x2=0

answer is D.

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Detailed Solution

Equation of AM=x−4+yα=1 Equation of BN = x4+y4+α=1 Eliminating α we get the locus of P as x2+2xy−16=0

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