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Chapter 6: Problem 45

Find the vertex, focus, and directrix of the parabola. Use a graphing utility to graph the parabola. $$y^{2}+x+y=0$$

Short Answer

Expert verified
The vertex of the parabola is at (0, -0.5). The focus of the parabola is at (0.25, -0.5). The directrix of the parabola is the vertical line x = -0.25. Graphing the equation \(y^{2} + x + y = 0\) confirms these findings.

Step by step solution

01

Rewrite equation in standard form

The standard form of a parabola equation that opens left or right is \((y - k)^2 = 4p(x - h)\), where (h, k) are the coordinates of the vertex and p is the distance from the vertex to the focus and the directrix. To get this, first rearrange the given equation to the form \(y^{2} = -x - y\), which can be rewritten in the standard form as \((y - (-0.5))^2 = - 4 * 0.25 * x\). Therefore, h = 0, k = -0.5 and p = -0.25.
02

Find the vertex

The vertex is given by the values of h and k from the standard form, therefore the vertex of the given parabola is at (0, -0.5).
03

Find the focus

For a parabola that opens left, the focus is at (h - p, k). Therefore the focus is at (0 - (-0.25), -0.5) = (0.25, -0.5).
04

Find the directrix

For a parabola that opens left, the directrix is a vertical line given by the equation x = h + p. Therefore, the directrix of the parabola is x = 0 + (-0.25) = -0.25.
05

Graph the parabola

Graphing the equation \(y^{2} + x + y = 0\) should result in a parabola opening to the left, with vertex at (0, -0.5), focus at (0.25, -0.5), and a vertical directrix at x = -0.25.

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