📂Geometry

Derivation of the Equation of an Ellipse

Formulas

The equation of an ellipse with the center at $(x_{0},y_{0})$, major axis $a$, and minor axis $b$ is as follows.

$$\frac{(x-x_{0})^{2}}{a^{2}}+\frac{(y-y_{0})^{2}}{b^{2}}=1$$

Description

An ellipse is a set of points where the sum of the distances to two foci is constant.

Derivation

Let’s consider an ellipse as shown in the figure above. Based on the definition of an ellipse, we can establish the following equation.

$$ \begin{align*} \overline{F^{\prime}P} +\overline{PF} =&\ \text{constant} \\ \sqrt{(x+c)^{2}+y^{2}}+\sqrt{(x-c)^{2}+y^{2}}=&
\end{align*} $$

When point $P$ is at $A$, we can understand that the constant sum of distances is $2a$. Therefore,

$$\sqrt{(x+c)^{2}+y^{2}}+\sqrt{(x-c)^{2}+y^{2}}=2a$$

Moving the first term on the left-hand side to the right side and squaring both sides gives us the following.

$$(x-c)^{2} + y^{2}=4a^{2}-4a\sqrt{(x+c)^{2}+y^{2}}+(x+c)^{2}+y^{2}$$

Now, by leaving only terms with a root on one side and rearranging, we get the following.

$$a\sqrt{(x+c)^{2}+y^{2}}=cx+a^{2}$$

Squaring both sides again, we get the following.

$$\begin{equation} \begin{align*} && a^{2}({\color{green}x^{2}} + 2cx + {\color{blue}c^{2}})+a^{2}y^{2} =&\ {\color{green}c^{2}x^{2}} + 2a^{2}cx + {\color{blue}a^{4}} \\ \implies&& {\color{green}(a^{2}-c^{2})x^{2}} + a^{2}y^{2}= & {\color{blue}a^{2}(a^{2}-c^{2})} \end{align*} \end{equation}$$

When point $P$ is at the position of $B$, by substituting $x=0$ and $y=b$ into the above equation, we get the following equation.

$$\begin{equation} \begin{align*} && a^{2}b^{2} =&\ a^{2}(a^{2}-c^{2}) \\ \implies && b^{2}=&\a^{2}-c^{2} \end{align*} \end{equation}$$

Substituting $(2)$ back into $(1)$ gives us the following equation.

$$\begin{align*} && b^{2}x^{2}+a^{2}y^{2} =&\ a^{2}b^{2} \\ \implies && \frac{x^{2}}{a^{2}}+\frac{y^{2} }{b^{2}} =&\ 1 \end{align*}$$

If the center of the ellipse is at $(x_{0},y_{0})$, then moving all points of the ellipse centered at the origin by $x_{0}$ along the $x$ axis and by $y_{0}$ along the $y$ axis is equivalent to

$$\frac{(x-x_{0})^{2}}{a^{2}}+\frac{(y-y_{0})^{2} }{b^{2}}=1$$

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