Solving for Parabolas?

Question

Specify the focus, the directrix, and the focal width as well as the vertex of the following parabola


y + square root 2 = [x-(2square root 2)]^2


any and all help is very much appreciated...thank you =)

Answer 1

4(1/4)(y + sqrt 2) = [x-(2sqrt 2)]^2

This is of the form 4py=+x^2: U,opening up

vertex is (2sqrt 2, -sqrt 2)

focus is a distance (1/4) from vertex, namely (2sqrt 2, 1/4 -sqrt 2)

directrix is a distance (1.4) from vertex, namely
y = -1/4 -sqrt 2

Answer 2

4p = 1 => p = 1/4, the distance from the vertex to the focus
Since the parabolas opens up, the directrix is
y = -√2 - p = -√2 - (1/4)
vertex: (2√2, -√2)
focus: [2√2, -√2 + p] = [2√2, -√2 + (1/4)]

Answer 3

the focus are the coordinance of the lowest point on the parabola, the directrix is the line under the parabola (forgot how to do it, i did alg2 last year in 11th but i forgot it all :]) there is a formula too...tough stuff i hated it!!!

Answer 4

y + square root 2 = [x-(2square root 2)]^2
y + sqrt(2) = x^2 -4xsqrt(2) + 8
y = x^2 -4xsqrt(2) +8 - sqrt(2)
So a = 1, b= -4sqrt(2), and c = 8-sqrt(2)
Now use quadratic formula:
x = [-b +/- sqrt(b^2-4ac)]/(2a)
x = [4sqrt(2) +/- sqrt(32 - 4 (1)(8-sqrt(2)]/2
x= [ 4sqrt(2) +/- sqrt(32 -32 +4sqrt(2)]/2
x = 2sqrt(2) +/- sqrt(4sqrt(2))]/2
x = sqrt(2) +/- sqrt(sqrt(2))
x = sqrt(2) +/- 2^.25

Answer 5

y+sqrt2 = (x-2sqrt2)^2 ?
1(y+sqrt2) = (x -2sqrt2)^2
vetex: (2sqrt2, -sqrt2) or (2.83, -1.4)
4a= 1
a= 1/4
Focus is k + a
-1.4 + 1/4 = 1.15
directrix is k - a
1.4 - 1/4
D: x = -1.65
width of the focal is 1