# How do you find the center, vertices, foci and asymptotes of  x^2/7 - y^2/9=1?

Dec 21, 2017

The graph should look like this:
graph{x^2/7-y^2/9=1 [-10, 10, -5, 5]}

#### Explanation:

${x}^{2} / 7 - {y}^{2} / 9 = 1$

Since we are subtracting, we know that this is a hyperbola. Also, since the fraction with ${x}^{2}$ is positive, this hyperbola opens to the right and the left.
Therefore, equation is in the form ${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2}$
We know that:
$h = 0$
$k = 0$
$a = \sqrt{7}$
$b = \sqrt{9}$ or $3$.
$c = \sqrt{{a}^{2} + {b}^{2}}$ or $4$.
Knowing this, we already know which equations to use:
The center is always $\left(h , k\right)$
The vertices are $\left(h + a , k\right)$ and $\left(h - a , k\right)$
The foci are $\left(h + c , k\right)$ and $\left(h - c , k\right)$
Since this hyperbola opens sideways, the asymptotes are found by this formula:
$y = k \pm \frac{b}{a} \left(x - h\right)$

Therefore, we know that the center is $\left(0 , 0\right)$
The vertices are approximately $\left(2.645 , 0\right)$ and $\left(- 2.645 , 0\right)$
The foci are $\left(4 , 0\right)$ and $\left(- 4 , 0\right)$
The lines of asymptotes are $y = \frac{3 \sqrt{7}}{7} x$ and $y = - \frac{3 \sqrt{7}}{7} x$