If the focus of a parabola is (2,5) and the directrix is y=3, the equation of the parabola is
A
x2+4x+4y+20=0
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B
x2+4x−4y+20=0
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C
x2−4x−4y+20=0
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D
x2+4x+4y+10=0
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Solution
The correct option is Cx2−4x−4y+20=0 Let (x,y) be any point on the parabola then the distance of this point from focus is equal to distance from directrix.
Distance of point from focus =√(x−2)2+(y−5)2
Distance of point from directrix =|0.x+1.y−3|√12+02
∴√(x−2)2+(y−5)2=|y−3|
squaring both sides, we get ⇒(x−2)2+(y−5)2=(y−3)2 ⇒x2−4x+4+y2−10y+25=y2−6y+9 ⇒x2−4x−4y+20=0