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  x² = 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  x² =  -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

x² =  -4py is the equation of the parabola ⇒ x² =  -4(5)y = -20y

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

(5 - y) = x² + (y + 5)²

25 + y² - 10y = x² + y² + 10y + 25

The equation of the parabola is x² = -20y

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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 x² = -20y.