How do you graph 12y^2-25x^2=300 and identify the foci and asympototes?

Jan 4, 2018

The foci are $\left(0 , 3 \sqrt{3}\right)$ and $\left(0 , - 3 \sqrt{3}\right)$
The asymptotes are $y = \frac{5}{2 \sqrt{3}} \left(x\right)$ and $y = - \frac{5}{2 \sqrt{3}} \left(x\right)$

Explanation:

First, we see turn the equation so that it equals one.
$12 {y}^{2} - 25 {x}^{2} = 300 \implies \frac{12 {y}^{2}}{300} - \frac{25 {x}^{2}}{300} = 1$
You can simplify this to:
$\frac{{y}^{2}}{25} - \frac{{x}^{2}}{12} = 1$

Remember that a hyperbolic equation can be in two forms:
${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$ or ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2}$ and $c = \sqrt{{a}^{2} + {b}^{2}}$

Now, a graph of a hyperbola opens to the left and the right when ${x}^{2}$ is over ${a}^{2}$.
Also, a graph of a hyperbola opens to up and down when ${y}^{2}$ is over ${a}^{2}$.

Because of this, the foci of a hyperbola is $\left(h \pm c , k\right)$ when ${x}^{2}$ is over ${a}^{2}$.

Similarly, the foci of a hyperbola is $\left(h , k \pm c\right)$ when ${y}^{2}$ is over ${a}^{2}$.

The equation for the asymptote of a hyperbola is $k \pm \frac{b}{a} \left(x - h\right)$ when ${x}^{2}$ is over ${a}^{2}$.

Similarly, the equation for the asymptote of a hyperbola is $k \pm \frac{a}{b} \left(x - h\right)$ when ${y}^{2}$ is over ${a}^{2}$

Since ${y}^{2}$ is over ${a}^{2}$, we use our appropriate rules to get:
$c = \sqrt{12 + 25} \implies c = 3 \sqrt{3}$

Therefore, the foci are $\left(0 , 3 \sqrt{3}\right)$ and $\left(0 , - 3 \sqrt{3}\right)$
The asymptotes are $y = \frac{5}{2 \sqrt{3}} \left(x\right)$ and $y = - \frac{5}{2 \sqrt{3}} \left(x\right)$

I have a graph from desmos.com here to visualize it.