# Equation of the Ellipse

To find the equation of an ellipse, let $P\left( {x,y} \right)$ be any point of the ellipse and $M\left( {\frac{a}{e},y} \right)$ be the corresponding point on the directrix, as shown in the given diagram. Then by the definition of an ellipse, we have $\begin{gathered} \frac{{PF}}{{PM}} = e \\ \Rightarrow PF = ePM \\ \Rightarrow \sqrt {{{\left( {x – ae} \right)}^2} + {{\left( {y – 0} \right)}^2}} = e\sqrt {{{\left( {x – \frac{a}{e}} \right)}^2} + {{\left( {y – y} \right)}^2}} \\ \Rightarrow {\left( {x – ae} \right)^2} + {y^2} = {e^2}{\left( {x – \frac{a}{e}} \right)^2} \\ \Rightarrow {x^2} – 2aex + {a^2}{e^2} + {y^2} = {e^2}\left( {{x^2} – 2\frac{a}{e}x + \frac{{{a^2}}}{{{e^2}}}} \right) \\ \Rightarrow {x^2} – 2aex + {a^2}{e^2} + {y^2} = {e^2}{x^2} – 2aex + {a^2} \\ \Rightarrow {x^2} – {e^2}{x^2} + {y^2} = {a^2} – {a^2}{e^2} \\ \Rightarrow \left( {1 – {e^2}} \right){x^2} + {y^2} = {a^2}\left( {1 – {e^2}} \right) \\ \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2}\left( {1 – {e^2}} \right)}} = 1\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

It is clear from the given diagram that in the triangle $FOB$, we have the relation given as
$\begin{gathered} {a^2}{e^2} + {b^2} = {a^2} \\ \Rightarrow {a^2} – {a^2}{e^2} = {b^2} \\ \Rightarrow {a^2}\left( {1 – {e^2}} \right) = {b^2} \\ \end{gathered}$

Using this relation in equation (i), we have
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

This is the equation of the ellipse whose center is at origin and foci lie on the X-axis. The lengths of the semi-major and semi-minor axes of this ellipse are $a$ and $b$ respectively.

If the foci lie on the Y-axis, then the graph is as shown in the given diagram. In this case the equation of the ellipse will be
$\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$

NOTE: For this we use the relation ${a^2} – {b^2} = {a^2}{e^2}$