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Chapter 6: Problem 30

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$y=-2 x^{2}$$

Short Answer

Expert verified
The vertex of the parabola \(y = -2x^2\) is at the origin (0,0). Its focus is at point (0,1/8) and its directrix is the line \(y = -1/8\).

Step by step solution

01

Find the Vertex

The equation of the parabola is in the form \(y = ax^2\), the vertex form of a parabola is \(y = a(x - h)^2 + k\), where (h, k) is the vertex of the parabola. Comparing the given equation with the general one, we find that the given parabola has \(h = 0\) and \(k = 0\). Therefore, the vertex of the parabola is (0, 0).
02

Find the Focus

The focus of a downward parabola is given by \((h, k - 1/4a)\). So substituting \(h = 0\), \(k = 0\), and \(a = -2\), we find that the focus of the parabola is \((0, 0 - 1/(4*-2)) = (0, 1/8)\). Therefore, the focus of the parabola is (0, 1/8).
03

Find the Directrix

The directrix of a downward parabola is given by the line \(y = k + 1/4a\). So substituting \(k = 0\) and \(a = -2\), we find that the equation of the directrix is \(y = 0 + 1/(4*-2) = -1/8\). Therefore, the directrix of the parabola is the line \(y = -1/8\).
04

Sketch the Parabola

Now that we have the vertex, focus, and directrix, we can sketch the parabola. Remember that a negative 'a' means the parabola opens down. The vertex is at the point (0, 0), the focus is inside the parabola at (0, 1/8), and the directrix is the horizontal line \(y = -1/8\) which is outside the parabola

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