What are the focus and directrix of the parabola with the equation y = one-twelfthx2?

Answer 1

To find the focus and directrix of the given parabola, we need to first put it in standard form:

y = (1/12)x^2

Multiply both sides by 12 to eliminate the fraction:

12y = x^2

Now we can see that the parabola is symmetric with respect to the y-axis, and its vertex is located at the origin (0,0). The coefficient of x^2 is positive, so the parabola opens upwards.

To find the focus, we need to first calculate p, which is the distance from the vertex to the focus. For a parabola with equation y = 4px, the focus is located at (p,0). In our case, p = 1/(4a), where a is the coefficient of x^2. So:

a = 1/12
p = 1/(4a) = 3

Therefore, the focus is located at (3,0).

To find the directrix, we use the formula x = -p. In our case, this gives:

x = -3

So the directrix is a vertical line with equation x = -3.

Answer 2

AAAaannndd the bot gets it wrong yet again!

You know that the parabola
x^2 = 4py
has focus at (0,p) and directrix y = -p
So, since your equation is
y = 1/12 x^2, that is the same as
x^2 = 12y
So, p=3.
Now just finish it off