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A parabola has a Vertex of V = `(-2,3)` and a Focus F = `(-2,2(1)/2)` What is the...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 9, 2013 at 4:49 AM via web

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A parabola has a Vertex of V = `(-2,3)` and a Focus F = `(-2,2(1)/2)`

What is the equation?  

Determine the 4p value

Enter the equation in the form:

`(x-h)^2 = 4p(y-k)`    or  ` (y-k)^2 = 4p(x-h) `                  

Tagged with algebra2, math

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llltkl | College Teacher | Valedictorian

Posted July 9, 2013 at 5:39 AM (Answer #1)

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Since the x-coordinates of the vertex and focus are the same that is -2, this is a regular vertical parabola. So, the standard form of the equation of a parabola with a vertical axis and vertex at `(h, k) ` is given by:

`(x-h)^2=4p(y-k)`

Here, vertex`(h,k)=(-2,3)`

Focus`(h,k+p)=(-2,2(1)/2)`

So,` k+p=2(1)/2`

`rArr 3+p=2(1)/2`

`rArr p=5/2-3=-1/2`

Hence, `4p=4*-1/2=-2`

Since p is negative the parabola opens downward.

Now plugging the values of `(h,k)` and `4p` in the standard form of the equation of the parabola we get:

`(x+2)^2=-2(y-3)`

Therefore, the required equation of the parabola is `(x+2)^2=-2(y-3)` .

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