provided the emphasis and also directrix the a parabola , just how do we find the equation of the parabola?

If we think about only parabolas that open upwards or downwards, climate the directrix will be a horizontal heat of the type y = c .

let ( a , b ) be the focus and also let y = c it is in the directrix. Let ( x 0 , y 0 ) be any point on the parabola.

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any type of point, ( x 0 , y 0 ) on the parabola satisfies the definition of parabola, so there room two distances to calculate:

Distance in between the allude on the parabola come the focus Distance in between the suggest on the parabola come the directrix

To find the equation the the parabola, equate these two expressions and solve because that y 0 .

uncover the equation that the parabola in the instance above.

Distance in between the point ( x 0 , y 0 ) and also ( a , b ) :

( x 0 − a ) 2 + ( y 0 − b ) 2

distance between allude ( x 0 , y 0 ) and also the line y = c :

|   y 0 − c   |

(Here, the distance between the allude and horizontal heat is difference of their y -coordinates.)

Equate the two expressions.

( x 0 − a ) 2 + ( y 0 − b ) 2 = |   y 0 − c   |

Square both sides.

( x 0 − a ) 2 + ( y 0 − b ) 2 = ( y 0 − c ) 2

expand the expression in y 0 top top both sides and also simplify.

( x 0 − a ) 2 + b 2 − c 2 = 2 ( b − c ) y 0

This equation in ( x 0 , y 0 ) is true because that all other values top top the parabola and also hence we have the right to rewrite v ( x , y ) .

Therefore, the equation the the parabola with focus ( a , b ) and directrix y = c is

( x − a ) 2 + b 2 − c 2 = 2 ( b − c ) y




You are watching: Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1.

Example:

If the focus of a parabola is ( 2 , 5 ) and also the directrix is y = 3 , find the equation the the parabola.

let ( x 0 , y 0 ) be any point on the parabola. Discover the distance in between ( x 0 , y 0 ) and the focus. Then find the distance in between ( x 0 , y 0 ) and also directrix. Equate these 2 distance equations and the simplified equation in x 0 and y 0 is equation that the parabola.

The distance in between ( x 0 , y 0 ) and also ( 2 , 5 ) is ( x 0 − 2 ) 2 + ( y 0 − 5 ) 2

The distance between ( x 0 , y 0 ) and also the directrix, y = 3 is

|   y 0 − 3   | .

Equate the 2 distance expressions and square top top both sides.

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( x 0 − 2 ) 2 + ( y 0 − 5 ) 2 = |   y 0 − 3   |

( x 0 − 2 ) 2 + ( y 0 − 5 ) 2 = ( y 0 − 3 ) 2

Simplify and also bring all terms to one side:

x 0 2 − 4 x 0 − 4 y 0 + 20 = 0

create the equation through y 0 ~ above one side:

y 0 = x 0 2 4 − x 0 + 5

This equation in ( x 0 , y 0 ) is true because that all other values on the parabola and also hence we have the right to rewrite with ( x , y ) .

So, the equation the the parabola with emphasis ( 2 , 5 ) and also directrix is y = 3 is