# Thread: Graph the ellipse and graph the hyperbola

1. ## Graph the ellipse and graph the hyperbola

Graph the ellipse. 16x^2+4y^2=48

what is the center?
Where are the vertices?
Where are the co-vertices?
Where are the foci?

Graph the hyperbola. 16y^2-4x^2=64
What is the center?
What are the vertices?
Where are the co-vertices?
Where are the foci?
What are the asymptotes?

2. Originally Posted by smartcp
Graph the ellipse. 16x^2+4y^2=48

what is the center?
Where are the vertices?
Where are the co-vertices?
Where are the foci?
Most folks resort to a Standard Form, usually this one

$\frac{x^{2}}{3} + \frac{y^{2}}{12} = 1$

Using standard notation:

This makes $a = \sqrt{3}$, and $b = 2\sqrt{3}$

Find 'c' and you nearly are done.

Note: If you REALLY cannot find the center of this one, you MUST go back and reread a section or two. You missed something very important.

3. Originally Posted by smartcp

Graph the ellipse. 16x^2+4y^2=48

what is the center?
Where are the vertices?
Where are the co-vertices?
Where are the foci?

Graph the hyperbola. 16y^2-4x^2=64
What is the center?
What are the vertices?
Where are the co-vertices?
Where are the foci?
What are the asymptotes?
Graph of Ellipse:

I'll let you figure out the rest.

Graph of Hyperbola:

Try to figure out the rest.

4. Originally Posted by smartcp

Graph the ellipse. 16x^2+4y^2=48

what is the center?
Where are the vertices?
Where are the co-vertices?
Where are the foci?

Graph the hyperbola. 16y^2-4x^2=64
What is the center?
What are the vertices?
Where are the co-vertices?
Where are the foci?
What are the asymptotes?
1.)
$16x^2+4y^2=48$
$\frac{x^2}{3} + \frac{y^2}{12} = 1$

2.)
$16y^2-4x^2=64$
$\frac{y^2}{4} - \frac{x^2}{16}=1$

Standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Vertices: $(\pm a, 0)$
Co-vertices: $(0, \pm b)$
Eccentricity: $e<1, \ \ b^2 = a^2(1-e^2)$
Foci: $(\pm ae, 0)$