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Question

The equation of the parabola whose focus is at (1,2) and the directrix is the line x2y+3=0

A
4x2+y2+4xy+4x+32y+16=0
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B
4x2+y2+4xy+2x+8y+16=0
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C
4x2+y2+4xy+40x+16y+16=0
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D
4x2+4y2+4xy+4x+32y+16=0
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Solution

The correct option is A 4x2+y2+4xy+4x+32y+16=0
Let P(x,y) be any point on the parabola whose focus is S(1,2) and the directrix x2y+3=0.

Now, distance of P from focus = distance of P from directrix.
(x+1)2+(y+2)2=|x2y+3|1+4
(x+1)2+(y+2)2=(x2y+31+4)2
5[(x+1)2+(y+2)2]=(x2y+3)25(x2+y2+1+2x+4y+4)=(x2+4y2+94xy+6x12y)4x2+y2+4xy+4x+32y+16=0

This is the equation of the required parabola.

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