Find the equation for the parabola described it's vertex , directrix and focus . Vertex=(-e,-e) directrix=y= -2e focus=(-e,0)
Given: Vertex=(-e,-e), directrix: y= -2e, focus=(-e,0) (y-k) = (1/4a)(x-h)^2 Where (h,k) is the vertex, a is the distance from the vertex to the focus (and from the vertex to the directrix), and e is the distance from the vertex to the focus. h = -e k = -e a = e e = a/2 Therefore, substituting these values in the vertex form, we get: (y+e) = (1/4e)(x+e)^2 x^2 = 4ey x^2 = 2ay Therefore, the equation of the parabola is x^2 = 2ay.
Question
Updated 361 days ago|4/30/2023 12:15:45 AM
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Given: Vertex=(-e,-e), directrix: y= -2e, focus=(-e,0)

(y-k) = (1/4a)(x-h)^2

Where (h,k) is the vertex, a is the distance from the vertex to the focus (and from the vertex to the directrix), and e is the distance from the vertex to the focus.

h = -e
k = -e
a = e
e = a/2

Therefore, substituting these values in the vertex form, we get:

(y+e) = (1/4e)(x+e)^2

x^2 = 4ey

x^2 = 2ay

Therefore, the equation of the parabola is x^2 = 2ay.
Added 361 days ago|4/30/2023 12:15:45 AM

Questions asked by the same visitor
Allyii 11 months ago Find an equation for the parabola described by its vertex, directrix, and focus point: vertex (-e, -e); directrix y =-2e; focus (-e, 0) (y+e)^2 = 4e(x+e) (x-e)^2 = 4e(y-e) (x+e)^2 = 4e(y+e) (y-e)^2 = 4e(x-e)
Question
Updated 361 days ago|4/30/2023 12:16:06 AM

(y + e)^2 = 4e(x + e)

or

y^2 + 2ey + e^2 = 4ex + 4e^2

This is because the directrix is a horizontal line and the focus is on the negative side of the y-axis, which means the parabola opens to the left. Therefore, the equation should have the form (y + k)^2 = 4p(x - h), where (h, k) is the vertex and p is the distance between the vertex and the focus.
Added 361 days ago|4/30/2023 12:16:06 AM
Rated good by Aj25
Find an equation for the parabola described by its vertex, directrix, and focus point: vertex (-e, -e); directrix y =-2e; focus (-e, 0) (y+e)^2 = 4e(x+e) (x-e)^2 = 4e(y-e) (x+e)^2 = 4e(y+e) (y-e)^2 = 4e(x-e)
Question
Updated 334 days ago|5/27/2023 12:53:52 AM
(y + e)^2 = 4e(x + e)
or
y^2 + 2ey + e^2 = 4ex + 4e^2

This is because the directrix is a horizontal line and the focus is on the negative side of the y-axis, which means the parabola opens to the left. Therefore, the equation should have the form (y + k)^2 = 4p(x - h), where (h, k) is the vertex and p is the distance between the vertex and the focus.
Added 334 days ago|5/27/2023 12:53:47 AM
(y + e)^2 = 4e(x + e)
or
y^2 + 2ey + e^2 = 4ex + 4e^2

This is because the directrix is a horizontal line and the focus is on the negative side of the y-axis, which means the parabola opens to the left. Therefore, the equation should have the form (y + k)^2 = 4p(x - h), where (h, k) is the vertex and p is the distance between the vertex and the focus.
Added 334 days ago|5/27/2023 12:53:52 AM
Deleted by Aj25 [5/27/2023 12:53:54 AM], Unflagged by Aj25 [2/16/2024 2:18:29 AM]
Find the equation for the parabola described it's vertex , directrix and focus
Weegy: Hello how are you? User: Find the equation for the parabola described it's vertex , directrix and focus . Vertex=(-e,-e) directrix=y= -2e focus=(-e,0) (More)
Question
Updated 334 days ago|5/27/2023 12:54:39 AM
(x + e)^2 = 4e^2y + 4e^3. is the equation for the parabola described it's vertex , directrix and focus . Vertex=(-e,-e) directrix=y= -2e focus=(-e,0).
(x - h)^2 = 4p(y - k),
(x + e)^2 = 4p(y - k)
(x + e)^2 = 4p(y + p)
(x + e)^2 = 4e^2(y + e)
(x + e)^2 = 4e^2y + 4e^3

Thus, the equation for the parabola is:

(x + e)^2 = 4e^2y + 4e^3.
Added 334 days ago|5/27/2023 12:54:39 AM
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