    Question

# if all the chord of the curve 3x2−y2−2x+4y=0 which subtend a right angle at origin passes through fixed point (a,b). Then |a+b| is equal to:

A
0
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B
1
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C
2
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D
3
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Solution

## The correct option is B 1Given curve 3x2−y2−2x+4y=0⋯(i) Let the chord equation of the curve be lx+my=1⋯(ii) Homogenising (i) with the help of (ii) 3x2−y2−(2x−4y)(lx+my)=0 ⇒(3−2l)x2+(−1+4m)y2+(−2m+4l)xy=0 ∵ it subtends right angle at the origin ⇒(3−2l)+(−1+4m)=0 ⇒2m−l+1=0 ⇒l−2m=1⋯(iii) from (ii) and (iii), chord always passes through (1,−2) ⇒|a+b|=|−1|=1  Suggest Corrections  0      Similar questions
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