Find the focus and directrix of the parabola with the given equation.

x^2=8y

I got:
focus: (0, 2)
directrix: y = -2

y^2=12x

I got:
focus: (3, 0)
directrix: x = -3

Match the equation to the graph.

y^2=8x

I got:

x^2=8y

I got:

Find the standard form of the equation of the parabola using the information given.

Focus: (-15, 0); Directrix: x = 15

I got: y^2=-60x

Vertex: (8, -9); Focus: (8, -4)
I got: (x-8)^2=20(y+9)

Find the vertex, focus, and directrix of the parabola with the given equation.

(y-3)^2=4(x+2)

I got:
vertex: (2, -3)
focus: (3, -3)
directrix: x = 1

(x+2)^2=16(y-4)

I got:

vertex: (2, -4)
focus: (2, 0)
directrix: y = -8

2. Originally Posted by soly_sol

Find the focus and directrix of the parabola with the given equation.

x^2=8y

I got:
focus: (0, 2)
directrix: y = -2

y^2=12x

I got:
focus: (3, 0)
directrix: x = -3

Match the equation to the graph.

y^2=8x

I got:

x^2=8y

I got:

Find the standard form of the equation of the parabola using the information given.

Focus: (-15, 0); Directrix: x = 15

I got: y^2=-60x

Vertex: (8, -9); Focus: (8, -4)
I got: (x-8)^2=20(y+9)

Find the vertex, focus, and directrix of the parabola with the given equation.

(y-3)^2=4(x+2)

I got:
vertex: (2, -3)
focus: (3, -3)
directrix: x = 1

(x+2)^2=16(y-4)

I got:

vertex: (2, -4)
focus: (2, 0)
directrix: y = -8
good job, everything is just fine..

3. Originally Posted by soly_sol

...

Find the vertex, focus, and directrix of the parabola with the given equation.

I (y-3)^2=4(x+2)

I got:
vertex: (2, -3)
focus: (3, -3)
directrix: x = 1

II (x+2)^2=16(y-4)

I got:

vertex: (2, -4)
focus: (2, 0)
directrix: y = -8
Hello,

to I:

V(-2, 3), F(-1, 3), d: x = -3

to II:

V(-2, 4), F(-2, 8), d: y = 0

4. Originally Posted by earboth
Hello,

to I:

V(-2, 3), F(-1, 3), d: x = -3

to II:

V(-2, 4), F(-2, 8), d: y = 0
oh yeah, i dindn't notice all the way down. hehe