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

Jun 25, 2018

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

${\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

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

$\frac{36 {x}^{2}}{324} + \frac{9 {y}^{2}}{324} = \frac{324}{324}$
${x}^{2} / 9 + {y}^{2} / 36 = 1$

We can name some crucial values right from our equation. For instance, the center is $\left(0 , 0\right)$ 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:

${a}^{2} - {b}^{2} = {c}^{2}$
$36 - 9 = {c}^{2}$
$27 = {c}^{2}$
$3 \sqrt{3} = c$

Let's graph this! We can set the center at $\left(0 , 0\right)$. 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: $\left(0 , 0 \pm 6\right) = \left(0 , 6\right) , \left(0 , - 6\right)$

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: $\left(\pm 3 , 0\right) = \left(3 , 0\right) , \left(- 3 , 0\right)$

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

Foci: $\left(0 , 0 \pm 3 \sqrt{3}\right) = \left(0 , 3 \sqrt{3}\right) , \left(0 , - 3 \sqrt{3}\right)$

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]}