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

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by adithyaca32

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by adithyaca32

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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

simplify to get answer.

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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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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

simplify to get answer.

=======================================

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.

==================================