Answer:
Option B
Part a) The focus is [tex](1/28,0)[/tex]
Part b) The directrix is [tex]x=-1/28[/tex]
Part c) The equation is [tex]y^{2}= (1/7)x[/tex]
Step-by-step explanation:
step 1
Find the equation of the parabola
we know that
The parabola in the graph has a horizontal axis.
The standard form of the equation of the horizontal parabola is
[tex](y - k)^{2}= 4p(x - h)[/tex]
where
p≠ 0
The vertex of this parabola is at (h, k).
The focus is at (h + p, k).
The directrix is the line x= h- p.
The axis is the line y = k.
If p > 0, the parabola opens to the right, and if p < 0, the parabola opens to the left
In this problem we have that the vertex is the origin
so
(h,k)=(0,0)
substitute in the equation
[tex](y - 0)^{2}= 4p(x - 0)[/tex]
[tex]y^{2}= 4p(x)[/tex]
The points (7,1) and (7,-1) lies on the parabola-----> see the graph
substitute the value of x and the value of y in the equation and solve for p
[tex](1)^{2}= 4p(7)[/tex]
[tex]1= 28p[/tex]
[tex]p=1/28[/tex]
The equation of the horizontal parabola is
[tex]y^{2}= 4(1/28)(x)[/tex]
[tex]y^{2}= (1/7)x[/tex]
step 2
Find the focus
we know that
The focus is at (h + p, k)
Remember that
[tex](h,k)=(0,0)[/tex]
[tex]p=1/28[/tex]
substitute
[tex](0+1/28,0)[/tex]
therefore
The focus is at
[tex]F (1/28,0)[/tex]
step 3
Find the directrix
The directrix is the line x = h- p
Remember that
[tex](h,k)=(0,0)[/tex]
[tex]p=1/28[/tex]
substitute
[tex]x=0-1/28[/tex]
[tex]x=-1/28[/tex]
