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 [tex]y=\frac{-1}{12} x^{2}[/tex]
Solution:
Given that, parabola has a focus at (0, -3) and directrix at y = 3.
We have to find the parabola equation in standard form.
The standard form of parabola is given as:
[tex](x-h)^{2}=4 p(y-k)[/tex]
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)
[tex]\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}[/tex]
So option 3 is correct
