Respuesta :
A parabola of the form, x = (a)y² + (b)y + c has a vertex, (h, k)
where
k = -b/2a
and
h = (a)k² + (b)k + c
In your case:
k = -0/{2(3)}
k = 0
h = 3(0)²
h = 0
The vertex is: (0, 0)
Let f = the focal width
f = 1/4a
f = 1/12
The focus is at
(h + f, k)
(1/12, 0)
The equation of the directrix is:
x = h - f
x = -1/12
where
k = -b/2a
and
h = (a)k² + (b)k + c
In your case:
k = -0/{2(3)}
k = 0
h = 3(0)²
h = 0
The vertex is: (0, 0)
Let f = the focal width
f = 1/4a
f = 1/12
The focus is at
(h + f, k)
(1/12, 0)
The equation of the directrix is:
x = h - f
x = -1/12
The vertex will be [tex](0,0)[/tex] , Focus will be [tex](\frac{1}{12} ,0)[/tex] , Directrix will be [tex]-\frac{1}{12}[/tex] and focal width will be [tex](f)=\frac{1}{12}[/tex] of the given parabola.
What is Parabola ?
Parabola is a symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side.
Equation of parabola : [tex](x - h)^2 = 4a(y - k)[/tex]
The coordinates of the focus are[tex]:\ (a, 0)[/tex],
And the equation of the directrix, [tex]x =h-f[/tex]
We have,
[tex]x = 3y^2[/tex] [tex]....(i)[/tex]
Rewrite in vertex form,
[tex]x=3(y-k)^2+h[/tex]
Using Vertex form, [tex]x=a(y-k)^2+h[/tex]
Compare equation [tex](i)[/tex] with Vertex form,
We get,
[tex]a=3,\ \ h=0,\ \ k=0[/tex]
So, Vertex of the given expression are [tex](0,0)[/tex].
Now,
To find Focus i.e. ;
For this find distance [tex](f)[/tex] between vertex and focus of parabola,
i.e. Using the formula, [tex](f)=\frac{1}{4a}[/tex]
i.e. put value of [tex]a[/tex] in [tex]\frac{1}{4a}[/tex],
Distance [tex](f)=\frac{1}{4*3}=\frac{1}{12}[/tex]
Now,
Focus = [tex](h + f, k)[/tex]
[tex]=(0+\frac{1}{12} ,0)[/tex]
So, Focus [tex]=(\frac{1}{12} ,0)[/tex]
Now,
Find Directrix;
Using the given formula,
Directrix, [tex]x =h-f[/tex]
[tex]x =0-\frac{1}{12}[/tex]
[tex]x =-\frac{1}{12}[/tex]
So, Directrix of given parabola is [tex]-\frac{1}{12}[/tex] .
Hence, we can say that the vertex will be [tex](0,0)[/tex] , Focus will be [tex](\frac{1}{12} ,0)[/tex] , Directrix will be [tex]-\frac{1}{12}[/tex] and focal width will be [tex](f)=\frac{1}{12}[/tex] of the given parabola.
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