Answer :
Answer:
the vertex is at (0,0)
the focus is at (0, -10)
the directrix is y = 10
the focal width = |-40| = 40
Step-by-step explanation:
The equation of the parabola :
[tex]y =\frac{-1}{40}x^2[/tex]
Which shows that the given parabola is a vertical parabola and the standard form of the vertical parabola is written as :
[tex]y =\frac{1}{4p}(x-h)^2+k .........(1)[/tex]
the vertex is at (h,k)
the focus is at (h,k+p)
the directrix is y = k−p
the focal width = |4p|
So, changing the given equation of parabola into standard form we get :
[tex]y =\frac{1}{4\times -10}(x-0)^2+0[/tex]
On comparing it with equation (1) we get
p = -10 , h = 0 , k = 0
Therefore, the vertex is at (0,0)
the focus is at (0, -10)
the directrix is y = 10
the focal width = |-40| = 40
For a parabola with equation [tex]y=\frac{-1}{40} x^2[/tex]
The vertex = (0, 0)
Focus = (0, -10)
Directrix; y = 10
Focal width = 40
Properties of a parabolic equation
The given parabolic equation is:
[tex]y=\frac{-1}{40} x^2...........................(1)[/tex]
The general equation of a parabola is:
[tex]y=\frac{1}{4p} (x-h)^2+k...................(2)[/tex]
Comparing equations (1) and (2)
4p = -40
p = -40/4
p = -10
h = 0
k = 0
The vertex, (h, k) = (0, 0)
The directrix; y = k - p
y = 0 - (-10)
y = 10
Directrix; y = 10
Focus = (h, k+p)
Focus = (0, 0-10)
Focus = (0, -10)
Focal width = |4p|
Focal width = |-40|
Focal width = 40
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