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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+(y+2)2=32xy152
or 5[(x+1)2+(y+2)2]=3(4x2+y2+14xy+2y4x)
or 5[x2+1+2x+y2+4y+4]=12x2+3y2+312xy+6y12x
5x2+5+10x+5y2+20y+2012x23y23+12xy6y+12x=0
7x2+2y2+12xy+22x+14y+22=0
or 7x22y212xy22x14y22=0


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