A parabola with directrix at` x=a` implies that the parabola may opens up sideways towards to the left or right.
The position of the directrix with respect to the vertex point can be used to determine in which side the parabola opens.
If the directrix is on the left side of the vertex point then the parabola opens towards to the rights side.
If the directrix is on the right side of the vertex point...
A parabola with directrix at` x=a` implies that the parabola may opens up sideways towards to the left or right.
The position of the directrix with respect to the vertex point can be used to determine in which side the parabola opens.
If the directrix is on the left side of the vertex point then the parabola opens towards to the rights side.
If the directrix is on the right side of the vertex point then the parabola opens towards to the left side.
The parabola indicated in the problem has directrix of `x=-2` which is located on the left side of the the vertex `(0,0)` .
Thus, the parabola opens towards to the right side and follows the standard formula: `(y-k)^2=4p(x-h)` . We consider the following properties:
vertex as `(h,k)`
focus as `(h+p, k)`
directrix as `x=h-p`
Note: `p` is the distance of between focus and vertex or distance between directrix and vertex.
From the given vertex point `(0,0)` , we determine `h =0` and `k=0` .
Applying directrix `x=-2 ` and `k=0` on `x=h-p` we get:
`-2=0-p`
`-2=-p`
`(-1)*(-2)=(-1)*(-p)`
`2=p or p=2` .Plug-in the values: `h=0` ,`k=0` , and `p=2` on the standard formula, we get:
`(y-0)^2=4*2(x-0)`
`y^2=8x ` as the standard form of the equation of the parabola with vertex `(0,0)` and directrix `x=-2` .
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