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

Question

2184 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-02T14:41:35

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

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

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

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

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

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

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

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

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

The equation of an hyperbola centered in $C (x_{c}, y_{c})$ $C(x_c,y_c)$ with semi-axes $a$ $a$ and $b$ $b$ is:

$\frac{{(x - x_{c})}_{c}^{2}}{a^{2}} - \frac{{(y - y_{c})}_{c}^{2}}{b^{2}} = \pm 1$ $(x-x_c)^2/a^2-(y-y_c)^2/b^2=+-1$ .

So, our hyperbola is:

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

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

If the equation of a conic section is (x+6)281−(y+1)216=1(x+6)^2/81-(y+1)^2/16=1, how has its center been translated?