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# Find the vertex, focus, directrix, and focal width of the parabola. x^{2} = 28y

**Solution:**

Given, the equation of __parabola__ is x² = 28y ------------------- (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 = 28y

4p = 28

p = 28/4

So, p = 7

k + p = 0 + 7 = 7

Therefore, focus = (0, 7)

y = 0 - 7

y = -7

Therefore, the directrix is y = -7

|4p| = |4(7)|

= 28

Therefore, the focal width is 28

## Find the vertex, focus, directrix, and focal width of the parabola. x^{2} = 28y

**Summary:**

The vertex, focus, directrix, and focal width of the parabola x² = 28y are (0, 0), (0, 7), y = -7 and 28

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