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Question 475284: find the focus 1/4(x-2)^2=(y+3)
and find the vertex and Directrix of -1/8(y+3)^2=(x-1)
thank you
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the focus 1/4(x-2)^2=(y+3)
and find the vertex and Directrix of -1/8(y+3)^2=(x-1)
**
focus:
1/4(x-2)^2=(y+3)
(x-2)^2=4(y+3)
This is a parabola with axis of symmetry: x=2 of the standard form: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex. Parabola opens upwards.
For given equation:(x-2)^2=4(y+3)
vertex: (2,-3)
4p=4
p=1
focus: (2,-2) (one unit above the vertex on the axis of symmetry)
..
Vertex and Directrix:
-1/8(y+3)^2=(x-1)
-(y+3)^2=8(x-1)
This is a parabola with axis of symmetry: y=-3 of the standard form: (y-k)^2=4p(x-h), with (h,k) being the (x,y) coordinates of the vertex. Parabola opens leftward.
For given equation:-(y+3)^2=8(x-1)
vertex:(1,-3)
4p=8
p=4
Directrix: x=5 (a line 4 units to the right of the vertex and perpendicular to the axis of symmetry, y=-3)
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