Question

1. What are the focus and directrix of the parabola with the equation y= 1 12 x^ 2 ?

Answer 1

y = 1/12 x^2
x^2 = 12y = 4*3y
so
vertex at (0,0)
focus at (0,3)
directrix is y = -3

Answer 2

Oh no, even with oobleck's correction, the bot still got the directrix wrong.

If the focus is (0,3) for our given y = (1/12)x^2, then
the directrix is clearly y = -3

Answer 3

The focus and directrix of the parabola with the equation y= 1/12 x^2 are:

- Focus: (0, 3/8)
- Directrix: y = -3/8

To find the focus and directrix of a parabola in this form, we can use the formula:

- Focus: (0, 1/(4a))
- Directrix: y = -1/(4a)

Where "a" is the coefficient of the x^2 term in the equation (in this case, a = 1/12).

Plugging in the values, we get:

- Focus: (0, 1/(4(1/12))) = (0, 3/8)
- Directrix: y = -1/(4(1/12)) = -3/8

Answer 4

I apologize, I made an error in my previous response. The correct solution is:

y = 1/12 x^2

We can rewrite the equation in vertex form by completing the square:

y = 1/12 (x^2)

y = 1/12 (x^2 - 0) + 0

y = 1/12 (x^2 - 0) + 0^2

The vertex of the parabola is at (0, 0).

We can now find the focus and directrix using the formula:

- Focus: (0, 1/(4a))
- Directrix: y = -1/(4a)

where a = 1/12.

- Focus: (0, 1/(4(1/12))) = (0, 3)
- Directrix: y = -1/(4(1/12)) = -3/4

Therefore, the focus is at (0, 3) and the directrix is y = -3/4.

Answer 5

I apologize for the confusion, you are correct. The directrix of the parabola with equation y = (1/12)x^2 and focus (0,3) is y = -3. Thank you for bringing this to my attention.