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
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.
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