# What is the equation of a parabola having focus at (3,-4) and directrix x - y + 5 = 0?

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2014-11-24T07:42:27+05:30

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Focus = (3, -4)    directrix = x - y + 5 =0

distance of a point on parabola is same from the directrix and from focus.

the equation of a parabola with focus = (x1, y1) and directrix = (ax+by+c =0  is

(x - x1)² + (y-y1)² = (a x+ by + c)² / (a²+b²)

(x - 3)² + (y +4)² = ( x - y + 5)² / √2

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alternately,

equation of perpendicular to directrix =    y + x + c = 0
Let it pass through focus :    -4 + 3 + c = 0  => c = 1

axis of parabola is    y + x + 1 = 0    call this  Y = 0  --- equation 1

intersection of axis and directrix : (-3 , 2).
Midpoint of focus and intersection point = vertex of parabola = (0, -1).

line parallel to directrix and through vertex:    x - y + c = 0
0 - (-1)  + c = 0,  c = -1
x - y -1 = 0  ,  call this  X = 0    --- equation 2

distance between vertex and focus = a =  3√2

parabola  =    Y² = 4 a X    or  (x+y+1)² = 4 * 3√2 * (x-y-1)

(x + y +1)² = 12√2 (x-y-1)

you may expand and simplify.
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