# If the equation of a conic section is (x+6)^2/81-(y+1)^2/16=1, how has its center been translated?

Feb 2, 2015

This is the equation of an hyperbola centered in the point $C \left(- 6 , - 1\right)$, with semi-axes: $a = 9$ and $b = 4$.

The equation of an hyperbola centered in $O \left(0 , 0\right)$ with semi-axes $a$ and $b$ is:

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = \pm 1$.

If the second member is $1$ then the hyperbola has the two branches on the "left" and on the "right".

E.G.: ${x}^{2} / 4 - {y}^{2} / 9 = 1$

graph{x^2/4-y^2/9=1 [-10, 10, -5, 5]}

If the second member is $- 1$ then the hyperbola has the two branches "up and "down":

E.G.: ${x}^{2} / 4 - {y}^{2} / 9 = - 1$

graph{x^2/4-y^2/9=-1 [-10, 10, -5, 5]}

The equation of an hyperbola centered in $C \left({x}_{c} , {y}_{c}\right)$ with semi-axes $a$ and $b$ is:

${\left(x - {x}_{c}\right)}^{2} / {a}^{2} - {\left(y - {y}_{c}\right)}^{2} / {b}^{2} = \pm 1$.

So, our hyperbola is:

${\left(x + 6\right)}^{2} / 81 - {\left(y + 1\right)}^{2} / 16 = 1$.

graph{(x+6)^2/81-(y+1)^2/16=1 [-20, 10, -20, 20]}