*General 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.