what is the center and vertices of the hyperbola, 4y^2-16y-x^2+12=0 ?

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beckden | High School Teacher | (Level 1) Educator

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To find the center and the vertecies we can use the standard form `(y-a)^2/a^2 - (x-b)^2/b^2 = 1 ` of a hyperbola so we are going to use completing the square to turn our equation into standard form.

` 4(y^2 - 4y) - (x-0)^2 = -12`

 

`4(y^2 - 4y + 4 - 4) - (x - 0)^2 = -12`

`4((y - 2)^2 - 4) - (x - 0)^2 = -12`

`4(y-2)^2 - 16 - (x - 0)^2 = -12`

`4(y-2)^2 - (x - 0)^2 = 4`

`(y - 2)^2/1^2 - (x - 0)^2/4 = 1`

So the center is at (0, 2)

This is a hyperbola with it's transverse axis aligned with the y - axis the distance between the center and the verticies c is computed from the following formula:

`c^2 = a^2 + b^2` where `a^2 = 1` and `b^2 = 4` we get

` c = sqrt(5)`

 

So the center is at (0, 2) and the verticies are at `(0, 2+sqrt(5))` and `(0, 2-sqrt(5))` .

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