The general form of a parabola is x^2−6x−2y+7=0 .
What is the standard form of this parabola?
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(___)^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)
+128408 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)
+128408 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)
