How do you graph the ellipse #36x^2 + 9y^2 = 324#?

1 Answer
Jun 25, 2018

Answer:

The most efficient way is to simplify the equation, find crucial points, and plot them on the graph.

Explanation:

For future reference: Formula for equation of vertical hyperbola

#(x-h)^2/b^2+(y-k)^2/a^2=1#

We need this equation in a more simplified form since the standard form must be equal to 1:

#(36x^2)/324+(9y^2)/324=324/324#
#x^2/9+y^2/36=1#

We can name some crucial values right from our equation. For instance, the center is #(0,0)# since there aren't any #h# or #k# values. Our ellipse is vertical since the bigger denominator is under #y^2#. Our #a#, #b#, and #c# values can also be found. When we look back at our formula, we can tell that #a=6# and #b=3#. But how do we find #c#? We can find it using this:

#a^2-b^2=c^2#
#36-9=c^2#
#27=c^2#
#3sqrt3=c#

Let's graph this! We can set the center at #(0,0)#. You might know that the vertices will be #a# units away from the center in opposite directions. In our case, since the ellipse is vertical, they would be up and down the y-axis:

Vertices: #(0,0+-6) = (0,6), (0,-6)#

The covertices are #b# units away, but in the different "set" of directions (in this case, to the left and right of the vertex):

Covertices: #(+-3,0) = (3,0), (-3,0)#

The foci are on the same line, the major axis, but are #c# units away:

Foci: #(0,0+-3sqrt3) = (0,3sqrt3), (0,-3sqrt3)#

We can plot these points on a graph and try our best to draw a smooth line through the vertices and covertices. While the line doesn't pass through the foci, it's still an important part of the ellipse you need to know.

Here's a graph of the ellipse in case you're confused:

graph{x^2/9+y^2/36=1 [-20, 20, -10, 10]}