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Question

All the chords of the hyperbola 3x2−y2−2x+4y=0, subtending a right angle at the origin pass through the fixed point :

A
(1,2)
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B
(1,2)
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C
(1,2)
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D
None of these
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Solution

The correct option is A (1,2)
Let ax+by=1 be the chord ...(1)
Making the equation of hyperbola homogeneous using (1), we get 3x2y2+(2x+4y)(ax+by)=0
(32a)x2+(1+4b)y2+(2b+4a)xy=0
Since the angle subtended at the origin is a right angle,so, coefficient of x2+ coefficient of y2=0
(32a)+(1+4b)=0a=2b+1
The chords are (2b+1)x+by1=0
b(2+y)+(x1)=0, which clearly pass through the fixed point (1,2).

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