Find an equation of the porabola described. - Vertex(0,0);directrix x + 1 = 0
To find the equation of a parabola, we need the vertex and either the focus or the directrix. In this case, we have the vertex (0, 0) and the equation of the directrix, which is x + 1 = 0.
First, let's determine the focus of the parabola. The focus is equidistant from the vertex and the directrix. Since the directrix is a vertical line (x + 1 = 0), the focus will also have the same x-coordinate as the vertex but a different y-coordinate. Therefore, the focus is (0, y), where y is the distance from the vertex to the directrix.
The distance from the vertex to the directrix is the absolute value of the constant term in the equation of the directrix, which is 1. So, the y-coordinate of the focus is 1 unit away from the vertex, giving us (0, 1).
Now, we can write the equation of the parabola using the vertex and the focus. The general equation of a parabola with a vertical axis of symmetry is given by (x - h)^2 = 4p(y - k), where (h, k) is the vertex and p is the distance from the vertex to the focus. In our case, h = 0, k = 0, and p = 1.
Plugging in these values, we get (x - 0)^2 = 4(1)(y - 0), which simplifies to x^2 = 4y.
Therefore, the equation of the parabola described by the given vertex and directrix is x^2 = 4y.