# Equation of an ellipse

## General equation of an ellipse

If we want to set equation of an ellipse we are going to remind some facts about an ellipse.

Ellipse whose center is matching the origin of the coordinate system, direction of the major axis with the $x$-axis, and the direction of the minor axis with the $y$-axis is defined by the following equation:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Where $a$ and $b$ are the semi-major and semi-minor axis.

We can say:

“Every ellipse is uniquely defined by its axes and the coordinates of its focuses.”

First, we’ll set ellipse in a coordinate system in a way that the center of the ellipse falls into the origin, that the direction of the major axis falls on the $x$-axis, and minor axis falls on the $y$-axis. Using the distance formula we get to the following definition:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Example 1. Construct an ellipse with equation $\frac{x^2}{81} + \frac{y^2}{16} = 1$

From simple comparison of the general equation and the given equation we can see that $a = 9$ and $b = 4$. This means that the coordinates of the vertexes are an equation we can see that $a=9$ and $b=4$. the minor axis with the $y$-axis has $A(-9,0), B(9,0), C(0,-4), D(0,4)$.

For the construction we’ll also be needing the linear eccentricity.

$e^2 = a^2 + b^2 = 65$

$e = \sqrt{65} \approx 8.1$ Can a major axis of an ellipse be aligned with $y$ – axis?

What if the equation of the ellipse was set as $\frac{x^2}{16} + \frac{y^2}{81} = 1$?

Now, the major axis will be aligned with $y$-axis and the minor axis with $x$-axis. In that case the focal points of this ellipse are now on the $y$ axis. What about ellipses whose center isn’t in the origin?

Example 2. Construct an ellipse with equation $\frac{(x – 1)^2}{81} + \frac{(y + 2)^2}{16} = 1$.

First thing you notice here is that the center of this ellipse is not in the origin. Finding its center is the same as finding the center of a circle. This means that for the $x$ coordinate we observe the square involving the $x$. Since that expression is $(x – 1)^2$, the $x$ coordinate of the center is $1$ (because $1 – 1 = 0$). Equivalently, $y$ coordinate is $-2$. Now that we know the coordinates of its center, we have to mark its axis. In these cases axis will always be aligned with the $x$ and $y$ axis. After you’re done with drawing the axis, the only thing that is left to do is apply the lengths of axis and finish the sketch. 