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

If we think about only parabolas that open up upwards or downwards, climate the directrix will be a horizontal line the the form y = c .

allow ( a , b ) be the focus and also let y = c be the directrix. Let ( x 0 , y 0 ) be any allude on the parabola. any type of point, ( x 0 , y 0 ) top top the parabola satisfies the definition of parabola, therefore there space two ranges to calculate:

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

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

uncover the equation the the parabola in the instance above.

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

( x 0 − a ) 2 + ( y 0 − b ) 2

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

|   y 0 − c   |

(Here, the distance between the point and horizontal heat is difference of your 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

broaden the expression in y 0 top top both sides and 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 various other values on the parabola and also hence we deserve to rewrite through ( x , y ) .

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

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

You are watching: What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

Example:

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

let ( x 0 , y 0 ) it is in any point on the parabola. Find 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 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 two 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 come one side:

x 0 2 − 4 x 0 − 4 y 0 + 20 = 0

write the equation v y 0 top top one side:

y 0 = x 0 2 4 − x 0 + 5

This equation in ( x 0 , y 0 ) is true because that all various other values on the parabola and also hence we can rewrite through ( x , y ) .

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