# Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = -1.

**Solution:**

Focus = (0, 1)

Directrix of y = - 1

Consider a point (x, y) on the parabola

The distance of a point on the parabola from directrix is same to its distance from the focus

Distance of (x, y) from the focus (0, 1) is √[(x − 0)^{2} + (y - 1)^{2}]------->(1)

Distance of (x, y) from the directrix y = -1 is |y + 1|-----> (2)

So the equation will be from (1) and (2)

√[(x − 0)^{2} + (y - 1)^{2}] = |y + 1|

Square on both sides

(x - 0)^{2} + (y - 1)^{2} = (y + 1)^{2}

So we get

x^{2} + y^{2} - 2y + 1 = y^{2} + 2y + 1

x^{2} = 4y

Therefore, the equation of the parabola is x^{2} = 4y.

## Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = -1.

**Summary: **

The equation of the parabola with a focus at (0, 1) and a directrix of y = -1 is x^{2} = 4y.