# How do you graph 36x^2-4y^2=144 and identify the foci and asympototes?

Nov 10, 2016

The foci are F$= \left(\sqrt{40} , 0\right)$ and F'$\left(- \sqrt{40} , 0\right)$
The equations of the asymptotes are $y = 3 x$ and $y = - 3 x$

#### Explanation:

The equation represents a hyperbola.
${x}^{2} / 4 - {y}^{2} / 36 = 1$
The general equation is
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The center is $\left(0 , 0\right)$

The vertices are $\left(h + a , k\right)$ and $\left(h - a , k\right)$
The vertices are $\left(2 , 0\right)$ and $\left(- 2 , 0\right)$

The slope of the asymptotes are $\pm \frac{b}{a} = \pm \frac{6}{2} = \pm 3$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$
$y = 3 x$ and $y = - 3 x$
To determine the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$

$c = \pm \sqrt{4 + 36} = \pm \sqrt{40}$
the foci are $\left(h \pm c , k\right)$
the foci are F$= \left(\sqrt{40} , 0\right)$ and F'$\left(- \sqrt{40} , 0\right)$
graph{(x^2/4-y^2/36-1)(y+3x)(y-3x)=0 [-18.02, 18.03, -9, 9.02]}