Find focus and directrix of parabola HELPP PLEASE
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Find focus and directrix of parabola HELPP PLEASE

[From: ] [author: ] [Date: 12-04-24] [Hit: ]
You now have it in the form (y-k)^2 = 2p(x -h) where (h,k) is the vertex and p/2 is the distance from the vertex to the focus on one siide and the directrix on the other side. The axis is parallel to x-axis.So vertex is at (-1/5,Since axis of symmetry is y = 1, The directrix is x = -5/4 -1/5 = -1.......
2y^2 -4y-10x=0

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2(y^2 - 2y + 1) = 10x + 2
(y - 1)^2 = 5x + 1
(y - 1)^2 = 5(x + 1/5)

The parabola opens to the right.
The vertex is (-1/5, 1)
4p = 5, so the focal length is 5/4
The focus has x-coordinate -1/5 + 5/4 = (-4 + 25)/20, so the focus is (21/20, 1)
The directrix is x = (-4 - 25)/20 = -29/20

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Divide by 2 getting:
y^2-2y = 5x
Add 1 to eeach side getting:
y^2-2y+1 = 5x + 1
(y-1)^2 = 5x +1= 5(x+1/5)
You now have it in the form (y-k)^2 = 2p(x -h) where (h,k) is the vertex and p/2 is the distance from the vertex to the focus on one siide and the directrix on the other side. The axis is parallel to x-axis.
So vertex is at (-1/5, 1)

P/2 = 5/4
Since axis of symmetry is y = 1, The directrix is x = -5/4 -1/5 = -1.45
The focus is at ( 1.05,1)
The parabola opens to the left
1
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