1. ## Hyperbola graphing help

Need help graphing focus, vertices, and asymptotic.
Y^2/16 - x^2/6=1. ( sorry I Havn't figured out how to wrap text). I got a=4 b= 2.4 c= 4.7 and y=4/2.4x or y=-4/2.4x. Does that sound right?

2. ## Re: Hyperbola graphing help

I got vertices as (0,4) and (0,-4) and foci (0,4.7) and (0,-4.7). Sound right?

3. ## Re: Hyperbola graphing help

We are given:

$\displaystyle \frac{y^2}{16}-\frac{x^2}{6}=1$

or:

$\displaystyle \frac{y^2}{4^2}-\frac{x^2}{\sqrt{6}^2}=1$

The foci will lie along the $\displaystyle y$-axis. We compute $\displaystyle c$:

$\displaystyle c=\sqrt{16+6}=\sqrt{22}$

Thus, the foci are at $\displaystyle (0,\pm\sqrt{22})$

The vertices are at:

$\displaystyle (0\pm4)$

The asymptotes are found by solving:

$\displaystyle \frac{y^2}{4^2}-\frac{x^2}{\sqrt{6}^2}=0$

$\displaystyle y=\pm\frac{2\sqrt{6}}{3}x$

4. ## Re: Hyperbola graphing help

How do you find other points on the hyperbolas?

5. ## Re: Hyperbola graphing help

Originally Posted by Nfalconer
How do you find other points on the hyperbolas?
Well, any point $\displaystyle (x,y)$ that satisfies the equation of the hyperbola will be on the hyperbola. In the case of this hyperbola, we could solve for $\displaystyle y$ as follows:

$\displaystyle y=\pm4\sqrt{1+\frac{x^6}{6}}$

And so for any arbitrary $\displaystyle x$ the points $\displaystyle \left(x,\pm4\sqrt{1+\frac{x^6}{6}}\right)$ will be on the hyperbola.