Expert: Sherry Wallin - 10/30/2009

Question
I need help with this problem, please.  Thank you!

Find the vertex, the focus, and the directrix.  Then draw the graph.

y^2 = -2x
Cindy


Answer
First you need to determine what this is. Is this a hyperbola, an ellipse, or a parabola? Both hyperbolas and ellipses have two squared variables and parabolas have just one so this is a parabola but since it is y^2 and not x^2 this means the parabola is lying parallel to the xaxis instead of parallel to the y axis. They are treated much the same except x and y change roles. So first solve for x in terms of y: x = y^2/-2 =(-1/2)y^2 The vertex in the more familiar parabola (in ax^2+bx+c) has a vertex -b/2a, here the a's and b's are the same (ay^2+by+c) so -b/2a is 0/2(-1/2) = 0. Then you evaluate (-b/2a, f(-b/2a)) so you get the origin as your vertex (0,0). Your equation is y^2 = -4ax = (-1/2)x so -4a = -1/2 so solving for 'a' you get a = 1/8. Your focus is (-a,0)  so the focus is (-1/8,0). The directrix of this parabola is (a,0) = (1/8,0).This parabola opens to the left. You can graph it by utilizing the vertex (0,0) and knowing the line of symmetry is y = 0 so if you need to plot a few points in the first or second quadrant they will be mirrored in the 3rd or 4th quadrant respectively. Also the focus is always 'inside' the parabola and the directrix is 'outside' the parabola.

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