How do I graph the ellipse represented by #(x-6)^2/36+(y+4)^2/16=1#?

1 Answer
Sep 14, 2014

Some calculators and software or websites will graph this for you:
enter image source here

In order for you to graph this yourself on paper, you would need to know how to find the center and the horizontal and vertical radii for the ellipse.

#(x-h)^2/(a^2)+(y-k)^2/(b^2)=1# is the form that tells us this information. (h,k) is the center of your graph. In this case, (6, -4). You can even see in the image above that the form is written as #(y-(-4))^2# in order to maintain the subtraction form.

The #a^2# that is the x-denominator tells us how far right and left to move away from the center. In this example, 36 = #6^2#, so we move right and left 6 units from the center. This puts your endpoints at (12,-4) and (0,-4). You can call this your "diameter" since it is the longer total distance moved from center, otherwise known as the major axis.

The #b^2# under the y-term tells us how far up and down to move from the center. If you move up and down 4 units, your endpoints are (6,0) and (6, -8). This is your shorter distance, or the minor axis.

In conclusion, this ellipse is easy to graph because it is in a nice form to work with. Think of it as a distorted circle, so the radii for the horizontal direction and the vertical direction are different from each other.