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Question

P:x2−y2+2y−1=0
L:x+y=3
Combined equation of L and L' is:

A
x2y24x+2y+3=0
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B
x2y22x+4y3=0
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C
x2y2+2x4y3=0
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D
x2y2+4x2y+3=0
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Solution

The correct option is B x2y22x+4y3=0
x2y2+2y1=0
x2(y22y+1)=0
x2(y1)2=0
(x(y1))(x+y1)=0
(xy+1)(x+y1)=0
Thus the lines are
xy+1=0
x+y1=0
Adding both, we get
2x=0
x=0
Substitution in any one equation gives us
y=1.
Thus the point of intersection is (0,1).
Now
L:x+y=3
Since L is perpendicular to L
Therefore
L:xy=d
It passes through (0,1)
d=1
Hence, L:xy+1=0
Thus the combined equation of L and L' is
(xy+1)(x+y3)=0
x2y23x+3y+x+y3=0
x2y22x+4y3=0

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