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Question

Find the equation of the hyperbola whose one directrix is 2xy=1, the corresponding focus is (1,2) and eccentricity is 3

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Solution

Equation of directrix is 2xy1=0.

Let S be the corresponding focus, then S=(1,2).

Given e=3

Let P(x,y) be any point on the required hyperbola.

Let PM be the length of the perpendicular from P to the directrix.

Then,PSPM=e=3

PS2=3PM2

or (x+1)2+(y2)2=32xy152

or 5[(x+1)2+(y2)2]=3(4x2+y2+14xy+2y4x)

or 5[x2+1+2x+y24y+4]=12x2+3y2+312xy+6y12x

5x2+5+10x+5y220y+2012x23y23+12xy6y+12x=0

7x2+2y2+12xy+22x26y+22=0

or 7x22y212xy+2x+26y22=0

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