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Question

All concurrent chords to the curve 3x2y22x+4y=0 , which subtends right angle to the origin pass through (a,b) then |a|+|b| equals to

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Solution

Given curve is
3x2y22x+4y=0, let y=mx+c and the chords by homogenization, combined equation of lines passing through point intersection of chords and curve
3x2y22x[ymxc]+4y[ymxc]=0
Itsubtends90attheorigin
Coefficientofx2+coeff.ofy2=0
3+2mc+4c1=0mc+2c+1=0
c=(m+2)
Hence, y=mx+c can be converted to
y=mxm2
m(x1)+(y2)=0
Which is of the form L1+λL2=0
Pointofconcurrencywillbe(1,2)
|a|+|b|=3

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