Find the vertex, focus, directrix, and focal width of the parabola. x = 4y2. [Solved]

A Step-by-step rationalization: i simply took the edg quiz

A Step-by-step rationalization: i simply took the edg quiz

A Step-by-step rationalization: i simply took the edg quiz

A Step-by-step rationalization: i simply took the edg quiz

Solutions: 1. None
from the alternatives, it needs to be
Vertex:
(0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 10 2.C)
y equals detrimental 1 divided by 8 x squared 3.D)
x equals detrimental 1 divided by 32 y squared 4.C)
x2 = -5.3y 5.C)
Heart: (0, 0); Vertices: (0, -15), (0, 15); Foci: (0, -12), (0, 12) 6.None
from the alternatives, it needs to be
Heart:
(0, 0); Vertices: the purpose detrimental two sq. root two comma zero
and the purpose 2 sq. root two comma zero; Foci: Ordered pair
detrimental sq. root 6 comma zero and ordered pair sq. root 6
comma zero 7.B)
A vertical ellipse is proven on the coordinate aircraft centered on the
origin with vertices on the level zero comma seven and 0 comma
detrimental seven. The minor axis has endpoints at detrimental six comma
zero and 6 comma zero. 8.D)
x squared divided by 16 plus y squared divided by 25 equals 1 9.D)
Vertices: (1, -1), (-11, -1); Foci: (-15, -1), (5, -1)

Half 1) Possibility C) y = detrimental one divided by thirty six x Half 2) Possibility A) y = one divided by thirty six x Half 3) Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = detrimental one divided by sixteen; Focal width: 0.25 Half 4) Possibility C) x = one divided by twelve y Step-by-step rationalization: Half 1) Discover the usual type of the equation of the parabola with a spotlight at (0, -9) and a directrix y = 9. we all know that The vertex type of the equation of the vertical parabola is the same as the place Vertex -> (h,okay) Focus -> F(h,okay p) directrix > y=k-p we now have F(0,-9) so h=0 okay p=-9 > equation A y=9 so k-p=9 -> equation B Provides equation A and equation B okay p=-9 k-p=9 2k=0 okay=0 so Discover the worth of p 0 p=-9 p=-9 substitute within the equation Convert to straightforward kind isolate the variable y Half 2) Discover the usual type of the equation of the parabola with a vertex on the origin and a spotlight at (0, 9) we all know that The vertex type of the equation of the vertical parabola is the same as the place Vertex -> (h,okay) Focus -> F(h,okay p) directrix > y=k-p we now have Vertex (0,0) > h=0,okay=0 F(0,9) so okay p=9> 0 p=9 > p=9 substitute within the equation Convert to straightforward kind isolate the variable y Half 3) Discover the vertex, focus, directrix, and focal width of the parabola. x = 4y we all know that The vertex type of the equation of the horizontal parabola is the same as the place Vertex -> (h,okay) Focus -> F(h p,okay) directrix > x=h-p we now have > so Vertex (0,0) > h=0,okay=0 4p=1/4 > Focal width p=1/16 Focus F(0 1/16,0) -> F(1/16,0) directrix > x=0-1/16 > x=-1/16 subsequently Vertex: (0, 0); Focus: one divided by sixteen comma zero; Directrix: x = detrimental one divided by sixteen; Focal width: 0.25 Half 4) Discover the usual type of the equation of the parabola with a spotlight at (3, 0) and a directrix at x = -3. we all know that The vertex type of the equation of the horizontal parabola is the same as the place Vertex -> (h,okay) Focus -> F(h p,okay) directrix > x=h-p we now have Focus F(3,0) so h p=3 > equation A okay=0 directrix x=-3 so h-p=-3 > equation B Provides equation A and equation B and resolve for h h p=3 h-p=-3 2h=0 > h=0 Discover the worth of p h p=3 > 0 p=3 > p=3 substitute within the equation Convert to straightforward kind isolate the variable x

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choice:A Vertex: (0,0) Focus: (0,-4) Directrix: y=4 focal width: 16 Step-by-step rationalization: Half 1: were given a equation of a parabola as: on changing the equation of parabola into the usual type of: we now have vertex as (h,okay). so right here the equation is transformed as: we now have h=0,okay=0 and p= -4 so, we now have vertex as (0,0). now the main target is given as (h,okay p) therefore focus is:(0,-4) directrix is given by the components y=k-p Therefore, directrix is_y=4. focal width of a parabola is given by |4 p| Therefore, Focal width is: 16. Therefore, choice A is right.

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