If the roots of the equationx12−a2m2−2x1y1m+y12+b2=0 a>bare the slopes of two perpendicular lines intersecting at Px1,y1, then the locus of P is

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x2+y2=a2+b2

x2+y2=a2−b2

x2−y2=a2+b2

x2−y2=a2−b2

answer is B.

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Detailed Solution

Equation x12−a2m2−2x1y1m+y12+b2=0 has roots m1 and m2∴ m1m2=y12+b2x12−a2=−1∴ x2+y2=a2−b2This is the required locus.

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