Q.

The locus of point of intersection of the line y+mx=a2m2+b2 and my−x=a2+b2m2 is

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a

x2+y2=1a2+1b2

b

x2+y2=a2+b2

c

x2−y2=a2−b2

d

1x2+1y2=a2−b2

answer is B.

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Detailed Solution

Let the point of intersection of given two lines is P(h, k), which lies on both the lines. ∴ k+mh=a2m2+b2 and mk−h=a2+b2m2Squaring and adding, we get k+mh2+mk-h2=a2m2+b2a2+b2m21+m2k2+1+m2h2=a2m2+b2+a2+b2m2 1+m2k2+h2 =a2m2+1+b2m2+1∴  locus of h,kis x2+y2=a2+b2
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