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Question

The equation of the hyperbola whose directrix is 2x+y=1,corresponding focus is (1,1) and eccentricity 3, is given by

A
7x2+12xy2y22x+4y7=0
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B
2x2+12xy7y22x+14y7=0
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C
7x212xy+2y22x+14y22=0
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D
7x2+12xy2y22x14y22=0
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Solution

The correct option is D 7x2+12xy2y22x+4y7=0
LetP(x,y) is any point on the hyperbola.
given, focus of parabola is S(1,1).
equation of directrix is 2x+y=1
From P draw PM perpendicular to the directrix then PM=(2x+y1)/(2²+1²)=(2x+y1)/5
Also from the definition of the hyperbola, we have
SP/PM=eSP=ePM
(x1)²+(y1)²=3(2x+y1)/5
(x1)²+(y1)²=3(2x+y1)²/5
5[(x²2x+1)+(y²2y+1)]=3(4x²+y²+1+4xy4x2y)
5x²10x+5+5y²10y+5=12x²+3y²+3+12xy12x6y
7x²+2y²+12xy2x+4y7=0
hence, equation of hyperbola is 7x²2y²+12xy2x+4y7=0

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