find the general equation of a given vertex of a parabola (-2, 2) and the focus (-4/2, 2)
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From the vertex we know that x+2=a(y-2)2, and because the y coord of the vertex and the focus are the same, so the parabola lies on its side. a is a constant. The vertex is the midpoint of the line from the focus to the vertical directrix. I'm guessing that the focus is meant to be (-4,2), 2 units to the left of the vertex. Therefore the directrix is 2 units to the right of the vertex, making it x=0 (y-axis). All points on the parabola are equidistant from the directrix line and the focus. When y=0, x+2=4a, so x=4a-2, the point P(4a-2,0). This point was chosen because we can work out easily how far it is from the focus and directrix. Distance from the directrix (y-axis) is the x-coord: 4a-2. The distance from the focus is:

√((4a-2-(-4))2+22)=√((4a+2)2+4). These distances are the same, so:

√((4a+2)2+4)=4a-2. Squaring each side we get:

(4a+2)2+4=(4a-2)2, 16a2+16a+4+4=16a2-16a+4, 32a=-4, a=-⅛.

Therefore the equation of the parabola is x+2=-⅛(y-2)2, or 8x+(y-2)2+16=0.

by Top Rated User (1.2m points)

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