Option 3 : y = negative 1 divided by 12 x^2
The standard form of the equation of the parabola with a focus at (0, -3) and a directrix at y = 3 is ![y=\frac{-1}{12} x^{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-1%7D%7B12%7D%20x%5E%7B2%7D)
<u>Solution:</u>
Given that, parabola has a focus at (0, -3) and directrix at y = 3.
We have to find the parabola equation in standard form.
<em><u>The standard form of parabola is given as:</u></em>
![(x-h)^{2}=4 p(y-k)](https://tex.z-dn.net/?f=%28x-h%29%5E%7B2%7D%3D4%20p%28y-k%29)
Where the focus is (h, k + p) and the directrix is y = k - p
So, here, (h, k + p) = (0, -3)
And y = k – p
y = 3
By comparison, h = 0, k + p = - 3 ------ eqn (2)
k – p = 3 ------ eqn (3)
Add (2) and (3)
k + p = -3
k – p = 3
(+)---------------
2k = 0
k = 0
Then, from (2) 0 + p = - 3
So, h = 0, k = 0, and p = -3
Now, put these values in (1)
![\begin{array}{l}{(x-h)^{2}=4 p(y-k)} \\\\ {(x-0)^{2}=4(-3)(y-0)} \\\\ {x^{2}=-12 y} \\\\{y=\frac{-1}{12} x^{2}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%28x-h%29%5E%7B2%7D%3D4%20p%28y-k%29%7D%20%5C%5C%5C%5C%20%7B%28x-0%29%5E%7B2%7D%3D4%28-3%29%28y-0%29%7D%20%5C%5C%5C%5C%20%7Bx%5E%7B2%7D%3D-12%20y%7D%20%5C%5C%5C%5C%7By%3D%5Cfrac%7B-1%7D%7B12%7D%20x%5E%7B2%7D%7D%5Cend%7Barray%7D)
So option 3 is correct