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Question

Find the equation of a parabola with focus at (1,2) and directrix x2y+3=0

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Solution

The focus of the parabola is F(1,2) and directrix is the line x2y+3=0
Let P(x,y) be any point in the plane of focus and directrix, and MP be the perpendicular distance from P to the directrix, then P lies on the parabola iff FP=MP
(x+1)2+(y+2)2=|x2y+3|12+(2)2
(x+1)2+(y+2)2=|x2y+3|1+4
(x+1)2+(y+2)2=|x2y+3|5
5[(x+1)2+(y+2)2]=(x2y+3)2
5[x2+2x+1+y2+4y+4]=x2+4y2+94xy12y+6x
5[x2+2x+y2+4y+5]=x2+4y2+94xy12y+6x
5x2+10x+5y2+20y+25x24y29+4xy+12y6x=0
4x2+y2+4xy+4x+32y+16=0 is the required equation of the parabola.

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