how do i find a hyperbola from the following equation? x^2-3y^2+8x-6y+4=0 i also need help finding the center

Expert Answers

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You should group the terms of equation such that:

`(x^2 + 8x) - (3y^2 + 6y) + 4 = 0`

You should complete the squares in the groups created such that:

`(x^2 + 8x + 16 - 16) - 3(y^2 + 2y + 1 - 1) + 4 = 0`

`(x + 4)^2 - 3(y + 1)^2 - 16 + 3 + 4 = 0`

`(x + 4)^2 - 3(y + 1)^2 = 9`

You need to divide by 9 such that:

`(x + 4)^2/9 - (y + 1)^2/3 = 1`

Hence, evaluating the equation of hyperbola yields:

`(x + 4)^2/3^2 - (y + 1)^2/(sqrt 3)^2 = 1`

`(x + 4)^2/9 - (y + 1)^2/3 = 1`

You should remember the standard equation of a hyperbola centered at (h,k) such that:

`(x - h)^2/a^2 - (y - k)^2/b^2 = 1`

Comapring the equations yields that the given hyperbola is centered at `(-4,-1).`

Hence, evaluating the equation of hyperbola yields `(x + 4)^2/9 - (y + 1)^2/3 = 1`  and evaluating its center yields `(-4,-1).`

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