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Question

All chords of the curve 3x2y22x+4y=0 which subtend a right angle at the origin, pass through the fixed point (a,b) which is equivalent to

A
(1,2)
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B
(1,2)
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C
(2,1)
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D
(2,1)
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Solution

The correct option is D (1,2)
Let y=mn+c is chord of 3x2y22x+4y=0

ymnc=1
By homogenisation,

3x2y22x(ymnc)+4y(ymnc)=0

3cx2cy2+2mx22xy+4y24mny=0

(3c+2m)x2+(4c)y2(x+4m)xy=0

Given θ=90

tanθ=0
a+b=0

(3c+2m)+(4c)=0

2c+2m+4=0

2=m+c
y=2 and x=1

Hence, option B is the correct answer.

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