Question

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

Answer 1

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]}