# Find the standard form of the equation of the parabola with the given characteristics. Vertex: at origin; Focus (0, -5).

**Solution:**

A set that consists of all the points in a plane equidistant from a given fixed point and a given fixed line in the plane is a** **parabola. The fixed point is the focus of the parabola. The fixed-line is the directrix.

The y-axis is called the axis of the parabola and the point where a parabola crosses its axis is the vertex. The vertex of the parabola x^{2} = 4py lies at the origin. The positive number p is the parabola’s focal length.

If the parabola opens downwards, with its focus at (0, -p) and its directrix the line y = p then the equation of the parabola is x^{2} = -4py

Given the vertex is V = (0,0)

The focus is F = (0,-5)

We know that focus coordinates are (0, -p)

Comparing we get, p = 5

So the directrix is y-5=0 (or) y =5

x^{2} = -4py is the equation of the parabola ⇒ x^{2} = -4(5)y = -20y

Any point on the parabola is equidistant from the focus and the directrix

(5 - y) = x^{2 }+ (y + 5)^{2}

25 + y^{2 }- 10y = x^{2 }+ y^{2 }+ 10y + 25

The equation of the parabola is x^{2 }= -20y

## Find the standard form of the equation of the parabola with the given characteristics. Vertex: at origin; Focus (0,-5).

**Summary:**

The standard form of the equation of the parabola with the given characteristics. Vertex: at origin; Focus (0, -5) is x^{2} = -20y.

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