# Find the vertex, focus, directrix, and focal width of the parabola. x^{2} = 12y

**Solution:**

Given, the equation of __parabola__ is x² = 12y ------------------- (1)

We have to find the vertex, focus, __directrix__ and focal width of the parabola.

The general equation of the parabola is given by

(x - h)² = 4p(y - k) -------------------------------------- (2)

Where, (h, k) is the vertex

(h, k + p ) is the focus

y = k - p is the directrix

|4p| is the focal width.

On comparing (1) and (2),

k = 0

h = 0

Therefore, vertex = (0, 0)

4py = 12y

4p = 12

p = 12/4

So, p = 3

k + p = 0 + 3 = 3

Therefore, focus = (0, 3)

y = 0 - 3

y = -3

Therefore, the directrix is y = -3

|4p| = |4(3)|

= 12

Therefore, the focal width is 12.

## Find the vertex, focus, directrix, and focal width of the parabola. x^{2} = 12y

**Summary:**

The vertex, focus, directrix, and focal width of the parabola x² = 12y are (0, 0), (0, 3), y = -3 and 12.

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