given the emphasis and directrix that a parabola , just how do we uncover the equation that the parabola?

If we take into consideration only parabolas that open up upwards or downwards, climate the directrix will certainly be a horizontal line of the kind y = c .

allow ( a , b ) it is in the focus and let y = c it is in the directrix. Allow ( x 0 , y 0 ) it is in any suggest on the parabola.

any kind of point, ( x 0 , y 0 ) top top the parabola satisfies the meaning of parabola, therefore there space two distances to calculate:

Distance between the suggest on the parabola to the focus Distance between the point on the parabola to the directrix

To uncover the equation of the parabola, equate these 2 expressions and solve for y 0 .

find the equation of the parabola in the example above.

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

( x 0 − a ) 2 + ( y 0 − b ) 2

street between suggest ( x 0 , y 0 ) and the heat y = c :

|   y 0 − c   |

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

Equate the 2 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 ~ above 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 hence we can rewrite through ( x , y ) .

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

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

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Example:

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

permit ( x 0 , y 0 ) be any allude on the parabola. Find the distance in between ( x 0 , y 0 ) and also the focus. Then discover the distance in between ( x 0 , y 0 ) and also directrix. Equate these 2 distance equations and the streamlined equation in x 0 and y 0 is equation of 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 in between ( x 0 , y 0 ) and also the directrix, y = 3 is

|   y 0 − 3   | .

Equate the 2 distance expressions and square ~ above 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 come one side:

x 0 2 − 4 x 0 − 4 y 0 + 20 = 0

write the equation through y 0 on one side:

y 0 = x 0 2 4 − x 0 + 5

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

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