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Question 589429: Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and the directrix, and the direction of opening of the parabola with equation: x=-y^2-2y+9
So far I have:
y^2+2y+x=9
(y^2+2y+1)+x=9+1
(y+1)^2+x=10
Here I got stuck.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and the directrix, and the direction of opening of the parabola with equation: x=-y^2-2y+9
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A standard form of equation for a parabola which opens leftwards:
(y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
For given equation:
x=-y^2-2y+9
complete the square
x=-(y^2+2y+1)+9+1
x=-(y+1)^2+10
(y+1)^2=-(x-10)
vertex:(10,-1)
axis of symmetry: y=-1
4p=1
p=1/4
focus=(10-1/4,-1)=(10-.25,-1)=(9.75,-1)
directrix: x=10.25 ( p units from vertex on the axis of symmetry
Parabola opens leftwards.
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note: You were on the right track. All you needed to know was the standard form of the equation for this particular parabola. There are 3 other configurations you should learn.
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