# find the equation of the parabola with coordinates of the vertex (0,0) and equation of the axis x=0, and passing through the point (-8,2) ans: x^2=32y

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Since, the given equation of the axis is x=0 so the standard form of the equation of a parabola with vertex at (h, k) is given by:

`(x-h)^2=4p(y-k)` (where the axis is vertical)

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Using the coordinates of the given vertex (0,0) and the coordinates of the given point (-8,2) we get:

`(-8-0)^2=4p(2-0)`

`rArr 64=8p`

`rArr p=8`

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` `Now plugging in the value of p in the equation of parabola we get:

`x^2=4*8y`

`rArr x^2=32y`

**Therefore, the required equation of the parabola is **`x^2=32y.`

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