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The general form of a parabola is x^2−6x−2y+7=0 .

What is the standard form of this parabola?

Enter your answer by filling in the boxes.

 

(___)^2 = ___ (___)

 

 

The vertex of a parabola is (6,2) , and the equation of its directrix is y = 4.

What is the equation of this parabola in standard form?

 

a. (y-2)^2= 8(x-6)

b. (x−6)^2= −8(y−2)

c. (y−2)^2= −8(x−6)

d. (x−6)^2= 8(y−2)

 Dec 6, 2019
 #1
avatar+109239 
+1

First one  :

 

x^2 - 6x  -  2y  + 7   = 0        rearrange as

 

2y - 7  =   x^2  -  6x                  complete the square on  x

 

Take (1/2)  of 6  = 3....square it  = 9.....add to both sides

 

2y - 7+ 9    =  x^2  - 6x + 9            factor the  right side, simplify the left

 

2y + 2   =  ( x - 3)^2           r

 

(x - 3)^2  =  2y + 2

 

( x - 3)^2  = 2 ( y + 1)

 

 

cool cool cool

 Dec 6, 2019
 #2
avatar+109239 
+1

Second one

 

The vertex of a parabola is (6,2) , and the equation of its directrix is y = 4.

What is the equation of this parabola in standard form?

 

 

The directrix is above the vertex.....therefore.....this parabola  turns downward

a = the distance between  the vertex and the  directrix  = 2

And the vertex  = (h, k) = ( 6, 2)

 

So....the form is 

 

(x - h)^2  = -4a ( y - k) 

 

(x - 6)^2  =  -8 ( y - 2)  

 

 

 

cool cool cool

 Dec 6, 2019

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