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Question 755032: Write the standard form of a parabola with a vertex (-6, 4) and a directrix x=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Write the standard form of a parabola with a vertex (-6, 4) and a directrix x=1
Equation is that of a parabola which opens leftward:
Its basic form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
vertex: (-6/4)
axis of symmetry: y=4
p=8 (distance from directrix to vertex on the axis of symmetry)
4p=32
equation: (y-4)^2=-32(x+6)
standard (vertex) form:
(y-4)^2=-32(x+6)
(y-4)^2=-32x-192
32x=-(y-4)^2-192
x=-(1/32)(y-4)^2-6
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