offered the focus and directrix the a parabola , how do we discover the equation the the parabola?

If we consider only parabolas that open upwards or downwards, then the directrix will certainly be a horizontal line the the type y = c .

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

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any point, ( x 0 , y 0 ) ~ above the parabola satisfies the an interpretation of parabola, for this reason there are two ranges to calculate:

Distance between the allude on the parabola come the emphasis Distance between the suggest on the parabola come the directrix

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

discover the equation that the parabola in the instance above.

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

( x 0 − a ) 2 + ( y 0 − b ) 2

street between point ( x 0 , y 0 ) and the line y = c :

|   y 0 − c   |

(Here, the distance in between the allude and horizontal heat 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

increase 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 for all other values top top the parabola and hence we deserve to rewrite with ( x , y ) .

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

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




You are watching: Using a directrix of y = −3 and a focus of (2, 1), what quadratic function is created?

Example:

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

allow ( x 0 , y 0 ) be any allude on the parabola. Uncover the distance between ( x 0 , y 0 ) and also the focus. Then uncover the distance in between ( x 0 , y 0 ) and directrix. Equate these 2 distance equations and also the streamlined equation in x 0 and y 0 is equation of the parabola.

The distance in between ( x 0 , y 0 ) and ( 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 on 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 bring every terms to one side:

x 0 2 − 4 x 0 − 4 y 0 + 20 = 0

create the equation v 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 top top the parabola and hence we have the right to rewrite through ( x , y ) .

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